Additional Questions - Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Q: Additional Questions - Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 11th Chemistry Guide Samacheer Kalvi Solutions

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On April 24, 2024, 11:35 AM

Additional Questions Solved
I. Choose the correct answer from the following
Question 1.
Which one of the following is the standard for atomic mass?
(a) ${ }_1 \mathrm{H}^1$
(b) ${ }_6 6 \mathrm{C}^{12}$
(c) ${ }_6 \mathrm{C}^{14}$
(d) ${ }_8 \mathrm{O}^{16}$
Answer:
(b) ${ }_6 6 \mathrm{C}^{12}$
Hint:
Standard element used to determine atomic mass is $6 \mathrm{Cl}_2$
Question 2.
One mole of $\mathrm{CO}_2$ contains
(a) $6.023 \times 10^{23}$ atoms of $\mathrm{C}$

(b) $6.023 \times 10^{23}$ atoms of $\mathrm{O}$
(c) $18.1 \times 10^{23}$ molecules of $\mathrm{CO}_2$
(d) $3 \mathrm{~g}$ atoms of $\mathrm{CO}_2$
Answer:
(a) $6.023 \times 10^{23}$ atoms of $\mathrm{C}$
Hint:
One mole of any substance contains Arogadro number of atoms. In this carbon one mole is present and oxygen two atoms are present. So, $6.023 \times 10^{23}$ atoms of $\mathrm{C}$ is correct.
Question 3.
The largest number of molecules is in
(a) $54 \mathrm{~g}$ of nitrogen pent oxide
(b) $28 \mathrm{~g}$ of carbon dioxide
(c) $36 \mathrm{~g}$ of water
(d) $46 \mathrm{~g}$ of ethyl alcohol
Answer:
(c) $36 \mathrm{~g}$ of water
Hint:(a) $54 \mathrm{~g}$ of $\mathrm{N}_2 \mathrm{O}_5$
$\mathrm{N}_2 \mathrm{O}_5=$ Molecular mass $=28+80=108$
$108 \mathrm{~g}$ of $\mathrm{N}_2 \mathrm{O}_5$ contains $6.023 \times 10^{23}$ molecules.
$\therefore 54 \mathrm{~g}$ of $\mathrm{N}_2 \mathrm{O}_5$ will contain $\frac{6.023 \times 10^{23}}{108} \times 54=3.0115 \times 10^{23}$ molecules.
(b) $28 \mathrm{~g}$ of $\mathrm{CO}_2$
$\mathrm{CO}_2=$ Molecular mass $=12+32=44$
$44 \mathrm{~g}$ of $\mathrm{CO}_2$ contains $6.023 \times 10^{23}$ molecules.
$\therefore 28 \mathrm{~g}$ of $\mathrm{CO}_2$ will contain $\frac{6.023 \times 10^{23}}{44} \times 28=3.832 \times 10^{23}$ molecules.
(c) $36 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$
$\mathrm{H}_2 \mathrm{O}=$ Molecular mass $=2+16=18$

$18 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ contains $6.023 \times 10^{23}$ molecules.
$\therefore 36 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ will contain $\frac{6.023 \times 10^{23}}{18} \times 36=12.046 \times 10^{23}$ molecules.
(d) 46 g of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}=$ Molecular mass $=24+6+16=46$
$46 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ contains $6.023 \times 10^{23}$ molecules.
So, among the $4,36 \mathrm{~g}$ of water contain the largest number of molecules as $12.046 \times 10^{23}$.
Question 4.
The number of moles of $\mathrm{H}_2$ in 0.224 liter of hydrogen gas at STP is
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(c) 0.01
22.4 liter of hydrogen gas at STP contains 1 mole.
$\therefore 0.224$ liter of hydrogen gas at STP will contain $\frac{1}{22.4} \times 0.224=0.01$

Question 5.
$10 \mathrm{~g}$ of hydrogen and $64 \mathrm{~g}$ of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be
(a) $3 \mathrm{~mol}$
(b) $4 \mathrm{~mol}$
(c) $1 \mathrm{~mol}$
(d) $2 \mathrm{~mol}$
Answer:
(b) $4 \mathrm{~mol}$
$10 \mathrm{~g}$ of $\mathrm{H}_2+64 \mathrm{~g}$ of $\mathrm{O}_2 \rightarrow$ water
$2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}$
$4 \mathrm{~g}$ of hydrogen react with $32 \mathrm{~g}$ of oxygen. So $10 \mathrm{~g}$ of hydrogen will react with $80 \mathrm{~g}$ of oxygen. But we have given amount of oxygen only $64 \mathrm{~g}$. It means, here oxygen is the limiting agent. Now all the oxygen react with $8 \mathrm{~g}$ of hydrogen and form 4 moles of water.
Question 6.
$6.023 \times 10^{20}$ molecules of urea are present in $100 \mathrm{ml}$ of its solution. The concentration of the solution is -
(a) $0.02 \mathrm{M}$
(b) $0.1 \mathrm{M}$
(c) $0.01 \mathrm{M}$
(d) $0.001 \mathrm{M}$
Answer:
(c) $0.01 \mathrm{M}$
$100 \mathrm{ml}$ of solution contain $6.023 \times 10^{20}$ molecules.
$6.023 \times 10^{23}$ molecules in $1000 \mathrm{ml}=1 \mathrm{M}$.

$
=10^{20} \times 10^{-23}=10^{-3}
$
$10^{-3}$ moles present in $100 \mathrm{ml}$
$\therefore$ In $1000 \mathrm{ml}$ the moles present is $=\frac{10^{-3}}{100} \times 1000=10^{-2}$ moles
Concentration $=0.01 \mathrm{M}$
Question 7.
Two containers $\mathrm{A}$ and 13 of equal volume contain $6 \mathrm{~g}$ of $\mathrm{O}_2$ and $\mathrm{SO}_2$ at $300 \mathrm{~K}$ and $1 \mathrm{~atm}$. Then
(a) No. of molecules in $\mathrm{A}$ is less than that in $B$
(b) No. of molecules in $A$ is more than that in $B$
(c) No. of molecules in A and B are same
(d) none of these
Answer:
(b) No, of molecules in A is more than that in $\mathrm{B}$
$\mathrm{O}_2=$ Mass $=6 \mathrm{~g}$
Molar mass $=32 \mathrm{~g}$
$32 \mathrm{~g}$ of $\mathrm{O}_2$ contains $6.023 \times 1023$ molecules.
$\therefore 6 \mathrm{~g}$ of $\mathrm{O}_2$ will contain $=6.023 \mathrm{X} 1023 \times 6=1.129 \times 10^{23}$ molecules.

$\mathrm{SO}_2=$ Mass $=6 \mathrm{~g}$
Molar mass $=64 \mathrm{~g}$
$64 \mathrm{~g}$ of $\mathrm{SO}_2$ contains $6.023 \times 10^{23}$ molecules.
$\therefore 6 \mathrm{~g}$ of $\mathrm{SO}_2$ will contain $6.023 \times 1023 \times 6=0.5646 \times 1023$ molecules.
$\therefore$ Number of molecules in $\mathrm{A}$ is more than that in $\mathrm{B}$.
Question 8.
The number of molecules in $16 \mathrm{~g}$ of methane is
(a) $3.023 \times 10^{23}$
(b) $6.023 \times 10^{23}$
(c) $16 / 6.023 \times 10^{23}$
(d) $6.023 / 3 \times 10^{23}$
Answer:
(b) $6.023 \times 10^{23}$
Hint:
Methane: $\mathrm{CH}_4$
Molecular mass $12+4=16$
$16 \mathrm{~g}$ of methane contains Avogadro number of molecules $6.023 \times 10^{23}$ molecules.
Question 9.
Number of atoms in $4.25 \mathrm{~g}$ of ammonia is
(a) $1 \times 10^{23}$
(b) $2 \times 10^{23}$
(e) $4 \times 10^{23}$
(d) $6 \times 10^{23}$
Answer:
(d) $6 \times 10^{23}$
Hint:
Ammonia $=\mathrm{NH}_3(4$ atoms)
Molecular mass $=14+3=17$

or
Density of water at $25^{\circ} \mathrm{C}=997.0479 \mathrm{~g} / \mathrm{L}$
Mass of $0.0018 \mathrm{ml}$ (or) $0.0018 \times 10^{-3} \mathrm{~L}$
$=\mathrm{D} \times \mathrm{V}$
$=997.05 \times 0.0018 \times 10^{-3}$
$=1.795 \times 10^{-3} \mathrm{~g}$
Molar mass of water $=18 \mathrm{~g}$
Mole $=\frac{\text { Mass }}{\text { Molecularmass }}=\frac{1.795 \times 10^{-3}}{18}$
$=9.971 \times 10^{-5}$
$\therefore$ Number of molecules in $0.0018 \mathrm{ml}=$ moles $\mathrm{x}$ Avogadro number
$=9.971 \times 10^{-5} \times 6.023 \times 10^{23}$
$=6 \times 10^{19}$ molecules.
Question 11.
$7.5 \mathrm{~g}$ of a gas occupy 5.6 liters of volume at STP. The gas
(a) $\mathrm{NO}$
(b) $\mathrm{N}_2 \mathrm{O}$
(c) $\mathrm{CO}$
(d) $\mathrm{CO}_2$
Answer:
(a) $\mathrm{NO}$
Hint:
22.4 liters $=1$ mole
5.6 liters $=\frac{1}{22.4} \times 5.6=0.25$ mole.
$\mathrm{NO}=$ molar mass $=14+16=30 \mathrm{~g}=1$ mole
$\mathrm{N}_2 \mathrm{O}=$ molar mass $=28+16=44 \mathrm{~g}=1$ mole
$\mathrm{CO}=$ molar mass $=12+16=28 \mathrm{~g}=1$ mole
$\mathrm{CO}_2=$ molar mass $=12+32=44 \mathrm{~g}=1$ mole
Among the four gases, 0.25 mole $=7.5 \mathrm{~g}$ is equal to $\mathrm{NO}$ gas.

Question 12.
The mass in grams of 0.45 mole of $\mathrm{CO}_2$ ions
(a) 1.8
(b) 40
(c) 36
(d) 18
Answer:
(d) 18
Hint:
$\mathrm{Ca}=$ Atomic mass $=40$
$\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}$
$41 \mathrm{~g}$ of $\mathrm{Ca}=1 \mathrm{~mole}$ (for $\mathrm{Ca}^{2+}$ Atomic mass remains same)
$1 \mathrm{~mole}$ of $\mathrm{Ca}^{2+}=40 \mathrm{~g}$
$\therefore 0.45$ mole of $\mathrm{Ca}^{2+}=\frac{40}{1} \times 0.45=18 \mathrm{~g}$

Question 13.
The mass of one molecule of $\mathrm{HI}$ in grams is
(a) $2.125 \times 10^{-22}$
(b) 128
(c) 127
(d) $6.02 \times 10^{-23}$
Answer:
(b) 128
Hint:
$\mathrm{HI}=1$ mole $=1+127=128 \mathrm{~g}$
Question 14.
Arogadro's number is the number of molecules present in
(a) $1 \mathrm{~g}$ of molecule
(b) $1 \mathrm{~g}$ atom of molecule
(c) gram molecular mass
(d) I lit of molecule
Answer:
(c) gram molecular mass
by definition (c) is correct
Question 15.
Which of the following contains same number of carbon atoms as are in $6.0 \mathrm{~g}$ of carbon (C-12)?
(a) $6.0 \mathrm{~g}$ ethane
(b) $8.0 \mathrm{~g}$ methane
(c) 21.0 g Propane
(d) $28.0 \mathrm{~g} \mathrm{CO}$
Answer:
(b) $8.0 \mathrm{~g}$ methane
Hint:
(a) 6.0 g of ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)$
$\mathrm{C}_2 \mathrm{H}_6=$ molar mass $=24+6=30 \mathrm{~g}$

$30 \mathrm{~g}$ of ethane contains $2 \times 6.023 \times 10^{23}$ Carbon atoms.
(b) $8.0 \mathrm{~g}$ of methane $\left(\mathrm{CH}_4\right)$
$\mathrm{CH}_4$ molar mass $=12+4=16 \mathrm{~g}$
$16 \mathrm{~g}$ of methane contains $6.023 \times 10^{23}$ Carbon atoms.
(c) $21.0 \mathrm{~g}$ of propane $\left(\mathrm{C}_3 \mathrm{H}_8\right)$
$\mathrm{C}_3 \mathrm{H}_8=$ molar mass $=36+8=44 \mathrm{~g}$
$44 \mathrm{~g}$ of propane contains $3 \times 6.023 \times 10^{23}$ Carbon atoms.
(d) $28.0 \mathrm{~g}$ of Carbon monoxide (CO)
$\mathrm{CO}=$ molar mass $=12+16=28 \mathrm{~g}$
$28 \mathrm{~g}$ of Carbon monoxide contains $6.023 \times 10^{23}$ Carbon atoms.
$6.0 \mathrm{~g}$ of Carbon contains $=6.023 \times 10 \times 6=3.0115 \times 10^{23}$ Carbon atoms.
Among the (a), (b), (c), (d) $-8 \mathrm{~g}$ of $\mathrm{CH}_4$ contains $\times 8=3.0115 \times 10^{23}$ Carbon atoms.

Question 16.
Equivalent mass of $\mathrm{KMnO}_4$ when it is converted to $\mathrm{MnSO}_4$ is equal to molar mass divided by
(a) 6
(b) 4
(c) 5
(d) 2
Answer:
(c) 5
Hint:
$
\begin{aligned}
& \stackrel{+7}{\mathrm{KMnO}_4} \rightarrow \mathrm{MnSO}_4^{+2} \\
& \mathrm{KMnO}_4+5 \mathrm{e}^{-} \rightarrow \mathrm{MnSO}_4
\end{aligned}
$
So, equivalent mass of $\mathrm{KMnO}_4=\frac{\text { Molar mass of } \mathrm{KMnO}_4}{5}$

Question 17.
How many equivalents of Sodium sulphate is formed when Sulphuric acid is completelyn neutralized with a base $\mathrm{NaOH}$ ?
(a) 0.2
(b) 2
(e) 0.1
(d) 1
Answer:
(d) 1
Hint:

Question 18.
$\mathrm{Cl}_2$ changes to $\mathrm{Cl}^{-}$and $\mathrm{ClO}^{-}$in cold $\mathrm{NaOH}$. Equivalent mass of $\mathrm{Cl}_2$ will be
(a) Molar mass / 2
(b) Molar mass / 1
(c) Molar mass / 3
(d) $2 \times$ Molar mass $/ 2$
Answer:

(a) Molar mass / 2

Question 19.
Equivalent mass of $\mathrm{KMnO}_4$ in acidic medium, concentrated alkaline medium and dilute basic medium respectively are $\mathrm{M}, \mathrm{M}, \mathrm{M}$. Reduced products can be
(a) $\mathrm{MnO}_2, \mathrm{MnO}_2^{2-}, \mathrm{Mn}^{2+}$
(b) $\mathrm{MnO}_2, \mathrm{Mn}^{2+}, \mathrm{MnO}_4^{2-}$
(c) $\mathrm{Mn}^{2+}, \mathrm{MnO}_2, \mathrm{MnO}_4^{2-}$
(d) $\mathrm{Mn}^{2+}, \mathrm{MnO}_4{ }^{2-}, \mathrm{MnO}_2$
Answer:
(c) $\mathrm{Mn}^{2+}, \mathrm{MnO}_2, \mathrm{MnO}_4{ }^{2-}$
Hint:
$
\begin{aligned}
& \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \text { (acidic medium) } \\
& \mathrm{MnO}_4+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O} \text { (concentrated basic medium) } \\
& \mathrm{MnO}_4+\mathrm{e}^{-} \rightarrow \mathrm{MnO}_4^{2-} \text { (dilute basic medium) }
\end{aligned}
$

Question 20.
The empirical formula of hydrogen peroxide is
(a) $\mathrm{HO}$
(b) $\mathrm{H}_2 \mathrm{O}$
(e) $\mathrm{H}_3 \mathrm{O}$
(d) $\mathrm{H}_2 \mathrm{O}_2$
Answer:
(a) $\mathrm{HO}$
Hint:
Molecular formula of hydrogen peroxide $=\mathrm{H}_2 \mathrm{O}_2$
$\mathrm{H}_2 \mathrm{O}_2 \div 2=\mathrm{HO}=$ Empirical formula .
Question 21.
Molecular mass $=$
(a) Vapour Density $\times 2$
(b) Vapour Density $\div 2$
(c) Vapour Density $\times 3$
(d) Vapour Density
Answer:
(a) Vapour Density $\times 2$
Question 22.
$20.0 \mathrm{~g}$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 \mathrm{~g}$ magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample?
(a) 60
(b) 84
(e) 75
(d) 96
Answer:
(b) 84
Hint:

$84 \mathrm{~g}$ of $\mathrm{MgCO}_3$ gives $40 \mathrm{~g}$ of $\mathrm{MgO}$. (100\% purity)
$20 \mathrm{~g}$ of $\mathrm{MgCO}_3$ will give $=\frac{40 \times 20}{84}=9.52 \mathrm{~g}$
$9.52 \mathrm{~g}$ of $\mathrm{MgO}$ is given by $100 \%$ pure $\mathrm{MgCO}_3$.
$8.0 \mathrm{~g}$ of $\mathrm{MgO}$ will be given by $=\frac{100 \times 8}{9.52}=84.03 \%$

Question 23.
What is the mass of the precipitate formed when the preparation of alkyl halides $50 \mathrm{ml}$ of $16.9 \%$ solution of $\mathrm{AgNO}_3$ is mixed with $50 \mathrm{ml}$ of $5.8 \% \mathrm{NaCl}$ solution?
(a) $7 g$
(b) $14 \mathrm{~g}$
(c) $28 \mathrm{~g}$
(d) $35 \mathrm{~g}$
Answer:
(a) $7 \mathrm{~g}$
Hint:

Question 24.
When $22 \mathrm{~L}$ of hydrogen gas is mixed with $11.2 \mathrm{~L}$ of chlorine gas, each at $\mathrm{STP}$, the moles of $\mathrm{HCl}$ gas formed is equal to ........
(a) 2
(b) 0.5
(c) 1.5
(d) 1
Answer:
(d) 1
Hint:
$
\mathrm{H}_2+\mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}
$
22 liters +11.2 litres
1 mole $+\frac{1}{2}$ mole $=1$ mole of $\mathrm{HCl}$ and $\frac{1}{2}$ mole of $\mathrm{H}_2$ is remained.
Question 25.
$5.6 \mathrm{~L}$ of a gas at STP are found to have mass of $11 \mathrm{~g}$. The molecular mass of the gas is
(a) 36
(b) 48
(c) 40
(d) 44
Answer:
(d) 44
Hint:
$5.6 \mathrm{~L}$ of gas has the mass of $11 \mathrm{~g}$.
$\therefore 22.4 \mathrm{~L}$ of gas will have the mass $\frac{11}{5.6} \times 22.4=44$
Question 26.
Oxidation number $f$ Fluorine in all compounds is
(a) +1
(b) -1

(c) 0
(d) -2
Answer:
(b) -1
Question 27.
In redox reaction which of the following is true?
(a) Number of electrons lost is more than number of electrons gained
(b) Number of electrons lost is less than number of electrons gained
(c) Number of electrons lost $s$ equal number of electrons gained
(d) No transfer and gain of electrons during the reaction.
Answer:
(c) Number of electrons lost is equal number of electrons gained
Question 28.
Which of the following is a mono-atomic molecule?
(a) Hydrogen
(b) Oxygen
(c) Sodium
(d) Ozone
Answer:
(c) Sodium
Question 29.
Which one of the following is a diatomic molecule?
(a) Ozone
(b) Copper
(c) Hydrogen
(d) Gold
Answer:
(c) Hydrogen

$
\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}=\text { Molecular mass }=(12 \times 2)+(1 \times 6)+(1 \times 16) \\
& =24+6+16=46
\end{aligned}
$
2 Carbon atoms are present.
$\therefore 2 \times 6.023 \times 10^{23} \mathrm{C}$ atoms is correct.
Question 32.
The mass of one mole of $\mathrm{CaCl}_2$ is
(a) $55.5 \mathrm{~g} \mathrm{~mol}^{-1}$
(b) $111 \mathrm{~g} \mathrm{~mol}^{-1}$
(c) $222 \mathrm{~g} \mathrm{~mol}^{-1}$
(d) $77.5 \mathrm{~g} \mathrm{~mol}^{-1}$
Answer:
(b) $111 \mathrm{~g} \mathrm{~mol}^{-1}$
Hint:
$
\mathrm{CaCl}_2=\text { Mass }=40+71=111 \mathrm{~g} \mathrm{~mol}^{-1}
$
Question 33.
$22 \mathrm{~g}$ of $\mathrm{CO}_2$ contains molecules of $\mathrm{CO}_2$
(a) $6.023 \times 10^{23}$
(b) $6.023 \times 10^{23}$
(c) $3.0115 \times 10^{23}$
(d) $3.0115 \times 10^{23}$
Answer:
(c) $3.0115 \times 10^{23}$
Hint:
$44 \mathrm{~g}$ of $\mathrm{CO}_2$ contains $6.023 \times 10^{23}$ molecules.
$\therefore 22 \mathrm{~g}$ of $\mathrm{CO}_2$ will contain $=\frac{6.023 \times 10^{23}}{44} \times 22=3.0115 \times 10^{23}$

Question 34.
The formula weight of ethanol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)$ is
(a) $56.5 \mathrm{amu}$
(b) $16 \mathrm{amu}$
(c) $60 \mathrm{amu}$
(d) $46 \mathrm{amu}$
Answer:
(d) 46 amu
Hint:
Formula weight of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}=(12 \times 2)+(6 \times 1)+(1 \times 16)$ $=24+6+16=46 \mathrm{amu}$
Question 35.
The number of moles of ethane in $60 \mathrm{~g}$ is .........
(a) 2
(b) 4
(c) 0.5
(d) 1
Answer:

(a) 2
Hint:
$\mathrm{C}_2 \mathrm{H}_6-$ Ethane - molar mass $=24+6=30 \mathrm{~g}$
$30 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_6$ contains $=1$ mole.
$\therefore 60 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_6$ will contains $=\mathrm{x} 1=2$ moles.
Question 36.
Which of the following method is used to prevent rusting of iron?
(a) Galvanization
(b) Painting
(c) Chrome plating
(d) all the above
Answer:
(d) all the above
Question 37.
Which of the following is not a redox reaction?
(a) $\mathrm{H}_2+\mathrm{F}_2 \rightarrow 2 \mathrm{HF}$
(b) $\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}$
(c) $2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}$
(d) $\mathrm{AgCl}+\mathrm{NH}_3 \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Cl}$
Answer:
(d) $\mathrm{AgCl}+\mathrm{NH}_3 \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Cl}$
Question 38.
How many $\mathrm{H}_2 \mathrm{O}$ molecules are there in a snowflake weighing $1 \mathrm{mg}$ ?
(a) $3.35 \times 10^{19}$
(b) $6.023 \times 10^{23}$
(c) $335 \times 10^{-19}$
(d) 100
Answer:

(a) $3.35 \times 10^{19}$
Hint:
$1 \mathrm{mg}$ of $\mathrm{H}_2 \mathrm{O}$
Molar mass of $\mathrm{H}_2 \mathrm{O}=2+16=18 \mathrm{~g}$
1 Mole contains $6.023 \times 10^{23}$ water molecules
$18 \mathrm{~g}$ contains $6.023 \times 10^{23}$ water molecules
$1 \mathrm{mg}$ contains $\frac{6.023 \times 10^{23}}{18} \times 1 \mathrm{~g} / 1000 \mathrm{mg}=3.35 \times 10^{-19} \mathrm{H}_2 \mathrm{O}$ molecule.
Question 39.
The volume of $\mathrm{HCl}$ gas weighing $73 \mathrm{~g}$ at STP is
(a) $2.24 \times 10^{-2} \mathrm{~m}^3$
(b) $4.48 \times 10^{-2} \mathrm{~m}^3$
(c) $4.48 \times 10^2 \mathrm{~m}^3$
(d) $2.24 \times 10^2 \mathrm{~m}^3$
Answer:
(b) $4.48 \times 10^{-2} \mathrm{~m}^3$

Hint:
$
\mathrm{HCl}=\text { Molar mass }=1+35.5=36.5 \mathrm{~g}
$
$36.5 \mathrm{~g}$ of $\mathrm{HCl}$ occupies $2.24 \times 10^{-2} \mathrm{~m}^3$
$\therefore 73 \mathrm{~g}$ of $\mathrm{HCl}$ at STP will occupy $\frac{2.24 \times 10^{-2}}{36.5} \times 73=4.48 \times 10^{-2} \mathrm{~m}^3$
Question 40 .
The molar volume of $22 \mathrm{~g}$ of $\mathrm{CO}_2$ is .........
(a) $2.24 \times 10^{-2} \mathrm{~m}^3$
(b) $4.48 \times 10^{-2} \mathrm{~m}^3$
(c) $1.12 \times 10^{-2} \mathrm{~m}^3$
(d) $2.24 \times 10^{-2} \mathrm{~m}^3$
Answer:
(c) $1.12 \times 10^{-2} \mathrm{~m}^3$
Hint:
$\mathrm{CO}_2=$ Molar'mass of $12+32=44 \mathrm{~g}$
$44 \mathrm{~g}$ of $\mathrm{CO}_2$ occupies molar volume $=2.24 \times 10^{-2} \mathrm{~m}^3$
$\therefore 2 \mathrm{~g}$ of $\mathrm{CO}_2$ will occupy $=\frac{2.24 \times 10^{-2}}{44} \times 2=1.12 \times 10^{-2} \mathrm{~L}$
Question 41.
The equivalent mass of Aluminium is
(a) 27
(b) 13.5
(c) 54
(d) 9
Answer:
(d) 9
Question 42.
The equivalent mass of $\mathrm{HSO}_4$ is ........
(a) 98
(b) 97

(c) 48
(d) 96
Answer:
(a) 98
Hint:
$\mathrm{HSO}_4=$ Molar mass $=1+32+64+1=98$
Equivalent mass $=$ Molar mass $/ 1=98$
Question 43.
The equivalent mass of $\mathrm{NaCl}$ is
(a) 40
(b) 58.5
(c) 35.5
(d) 23
Answer:
(b) 58.5
Hint:
$\mathrm{NaCl}=$ Salt Molar mass $23+35.5=58.5$
Equivalent mass of Salt = Molar mass of Salt.

Question 44.
Match the List-I and List-II using the correct code given below the list.

Answer:
(a) $4 \quad 3 \quad 2 \quad 1$
Question 45.
Match the List-I and List-Il using the correct code given below the list.

Answer:
(b) $3 \quad 4 \quad 2 \quad 1$
Question 46.
Consider the following statements
(i) Empirical formula shows the actual number of atoms of different elements in one molecule of the compound.
(ii) Ozone is a diatomic molecule.
(iii) Gases are easily compressible.
Which of the above statement is/are not correct?
(a) (i), (ii), (iii)
(b) (i) \& (ii)
(c) (ii) \& (iii)
(d) (iii) only
Answer:
(b) (i) & (ii)

Question 47.
How many molecules of hydrogen is required to produce 4 moles of ammonia?
(a) 15 moles
(b) 20 moles
(c) 6 moles
(d) 4 moles
Answer:
(c) 6 moles
Hint:
$
3 \mathrm{H}_2+\mathrm{N}_2 \rightarrow 2 \mathrm{NH}_3
$
To get 2 moles of ammonia, 3 mole of $\mathrm{H}_2$ is required.
To get 4 moles of ammonia $=\frac{3}{2}, \mathrm{x} 4$
$=6$ moles of $\mathrm{H}_2$ is required.
Question 48.
The number of moles of oxygen required to prepare 1 mole of water is
(a) I mole
(b) 0.5 mole
(c) 2 moles
(d) 0.4 mole
Answer:
(b) 0.5 mole
Hint:
$
\mathrm{H}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}
$
1 mole $\quad 0.5$ mole $1 \mathrm{~mole}$
0.5 mole of oxygen is required to prepare 1 mole of $\mathrm{H}_2 \mathrm{O}$.

Question 49.
How much volume of $\mathrm{CO}_2$ is produced when $50 \mathrm{~g}$ of $\mathrm{CaCO}_3$ is heated strongly?
(a) $2.24 \times 10^{-2} \mathrm{~m}^3$
(b) 22.4
(c) $11.2 \mathrm{~L}$
(d) $22400 \mathrm{~cm}^3$
Answer:
(c) $11.2 \mathrm{~L}$

$100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ produces 22.4 litres of $\mathrm{CO}_2$.
$50 \mathrm{~g}$ of $\mathrm{CaCO}_3$ will produce $=\frac{22.4}{100} \times 50=11.2$ litres
Question 50 .
Which one of the following is not a redox reaction?
(a) Rusting of iron
(b) Extraction of metal $\mathrm{Na}$
(c) Electroplating
(d) Aluminothermic process
Answer:
(a) Rusting of Iron
Question 51 .
In the reaction $2 \mathrm{AuCl}_3+3 \mathrm{SnCl}_2 \rightarrow 2 \mathrm{Au}+3 \mathrm{SnCl}_4$ which is an oxidising agent?
(a) $\mathrm{AuCl}_3$
(b) $\mathrm{Au}$
(c) $\mathrm{SnCl}_2$
(d) Both $\mathrm{AuCl}_3$ and $\mathrm{SnCl}_2$
Answer:
(a) $\mathrm{AuCl}_3$
Hint:
$\mathrm{AuCl}_3$ undergoes reduction. So, it is an oxidising agent.
Question 52 .
Identify the compound formed during the rusting of iron.
(a) $\mathrm{Fe}_2 \mathrm{O}_3$
(b) $\mathrm{Fe}_2 \mathrm{O}_3 \cdot \times \mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{FeO} \cdot \mathrm{x}_2 \mathrm{O}$
(d) $\mathrm{FeO}$
Answer:
(b) $\mathrm{Fe}_2 \mathrm{O}_3 \cdot \times \mathrm{H}_2 \mathrm{O}-$ Hydrated iron oxide is rust.

Question 53.
The oxidation state of a substance in its elementary state is equal to
(a) -1
(b) -2
(c) zero
(d) charge of the ion
Answer:
(c) zero
Question 54.
The oxidation number of fluorine in all its compounds is equal to
(a) -1
(b) +1
(c) -2
(d) +2
Answer:
(a) -1

Question 55.
Consider the following statements.
(i) The sum of the oxidation number of all the atoms in neutral molecule is equal to zero.
(ii) Fluorine has an oxidation number +1 in all its compounds. .
(iii) The oxidation number of a substance in its elementary state is equal to zero.
Which of the above statement is/are not correct?
(a) (i), (ii) & (iii)
(b) (ii) & (iii)
(c) (i) only
(d) (ii) only
Answer:
(d) (ii) only
Question 56.
The oxidation number of $\mathrm{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is
(a) +4
(b) +6
(c) $\mathrm{O}$
(d) +7
Answer:
(b) +6
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_2$
$2+2 \mathrm{x}-140$
$2 \mathrm{x}-12=0$
$2 \mathrm{x}=+12$
$\mathrm{x}=+6$
Question 57.
The oxidation number of $\mathrm{N}$ in $\mathrm{NH}_4$ ion is
(a) +4
(b) +3
(c) -3
(d) -4
Answer:
(c) -3
$\mathrm{NH}_2^{+}$
$
\begin{aligned}
& \mathrm{x}+4=+1 \\
& \mathrm{x}=+1-4 \\
& \mathrm{x}=-3
\end{aligned}
$
Question 58 .
$\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}^{2+}{ }_{(\mathrm{aq})} \rightarrow \mathrm{Zn}^{2+}{ }_{\text {(aq) }}+\mathrm{Cu}_{\text {(s) }}$. In this reaction, which gets oxidised?
(a) $\mathrm{Cu}^{2+}$
(b) $\mathrm{Zn}^{2+}$
(c) $\mathrm{Zn}$
(d) $\mathrm{Zn}, \mathrm{Cu}^{2+}$

Answer:
(c) $\mathrm{Zn}$
Hint:
$\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$(loss of electron = oxidation)
$\mathrm{Zn}$ gets oxidised
Question 59.
Which one of the following is an example for disproportionation reaction?
(a) $\mathrm{CuSO}_4+\mathrm{Zn} \rightarrow \mathrm{ZnSO}_4+\mathrm{Cu}$
(b) $2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCI}+3 \mathrm{O}_2$
(c) $\mathrm{PCl}_5 \rightarrow \mathrm{PCl}_3+\mathrm{Cl}_2$
(d) $4 \mathrm{H}_3 \mathrm{PO}_3 \rightarrow 3 \mathrm{H}_3 \mathrm{PO}_4+\mathrm{PH}_3$
Answer:
(d) $4 \mathrm{H}_3 \mathrm{PO}_3 \rightarrow 3 \mathrm{H}_3 \mathrm{PO}_4+\mathrm{PH}_3$ (Auto oxidation and reduction reaction)
Question 60 .
The number of molecules in $40 \mathrm{~g}$ of sodium hydroxide is .........
(a) $6.023 \times 10^{23}$
(b) $3.0115 \times 10^{23}$
(c) $6.023 \times 10^{23}$
(d) $2 \times 6.023 \times 10^{23}$
Answer:
(c) $6.023 \times 10^{23}$
Sodium hydroxide $=\mathrm{NaOH}=23+16+1=40 \mathrm{~g}$
$40 \mathrm{~g}=1 \mathrm{~mole}=6.02310^{23}$
Question 61 .
The mass of one molecule of $\mathrm{AgCl}$ in grams is
(a) $108 \mathrm{~g}$
(b) $143.5 \mathrm{~g}$
(c) $35.5 \mathrm{~g}$
(d) $243.5 \mathrm{~g}$
Answer:
(b) $143.5 \mathrm{~g}$

Hint:
Mass of $\mathrm{AgCl}=108+35.5=143.5 \mathrm{~g}$.
Question 62.
The empirical formula of Alkene is
(a) $\mathrm{CH}$
(b) $\mathrm{CH}_2$
(c) $\mathrm{CH}_3$
(d) $\mathrm{CH}_3 \mathrm{O}$
Answer:
(b) $\mathrm{CH}_2$
Hint:
Alkene $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}$ Molecular formula
E.F. $=$ M.F. $/ 2$
$\therefore$ Empirical formula $=\mathrm{CH}_2$
Question 63.
$22 \mathrm{~g}$ of a gas occupies 11.2 litres of volume at STP. The gas is
(a) $\mathrm{CH}_4$
(b) $\mathrm{NO}$
(c) $\mathrm{CO}$
(d) $\mathrm{CO}_2$
Answer:
(d) $\mathrm{CO}_2$
Hint:
$22 \mathrm{~g}$ of a gas occupies 11.2 litres.
11.2 liters is occupied by $22 \mathrm{~g}$ of a gas.
$\therefore$ Molar volume 22.4 liter will be occupied by $\frac{22}{11.2} \times 22.4=44 \mathrm{~g}$
$\therefore$ The gas is $\mathrm{CO}_2$.
Question 64.
The number of moles of $\mathrm{H}_2$ in 2.24 liter of hydrogen gas at STP is ........
(a) 1

(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(b) 0.1
Hint:
22.4 liter $=1$ molar volume $=1$ mole.
$\therefore 2.24$ liter $=\frac{1}{22.4} \times 2.24=01$ mole
Question 65.
How many molecules are present in $32 \mathrm{~g}$ of methane?
(a) $2 \times 6.023 \times 10^{23}$
(b) $6.023 \times 10^{23}$
(c) $6.023 \times 10^{23}$
(d) $3.011 \times 10^{23}$
Answer:
(a) $2 \times 6.023 \times 10^{23}$
Hint:
Methane $\left(\mathrm{CH}_4\right)-$ Molar mass $=12+4=16 \mathrm{~g}$.
$16 \mathrm{~g}$ contains $6.023 \times 10^{23}$ molecules.
$\therefore 32 \mathrm{~g}$ of methane will contain $=\frac{6.023 \times 10^{23}}{16} \times 32^2=2 \times 6.023 \times 10^{23}$
Question 66.
The empirical formula of glucose is
(a) $\mathrm{CH}$
(b) $\mathrm{CH}_2 \mathrm{O}$
(c) $\mathrm{CH}_2 \mathrm{O}_2$
(d) $\mathrm{CHO}$
Answer:
(b) $\mathrm{CH}_2 \mathrm{O}$
Hint:
Glucose $=$ Molecular formula $=\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
Empirical formula $=\frac{\text { Molecular formula }}{6}=f r a c \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 6=\mathrm{CH}_2 \mathrm{O}$

Question 67.
How many moles of water is present in I L of water?
(a) 1
(b) 18
(c) 55.55
(d) 5.555
Answer:
(c) 55.55
Hint:
1 Liter of water $=1000 \mathrm{~g}$.
Molar mass of water $18 \mathrm{~g}$
Number of moles $=\frac{\text { Mass }}{\text { Molarmass }}=\frac{1000}{18}=55.55$ moles
Question 68 .
How many moles of Hydrogen atoms are present in 1 mole of $\mathrm{C}_2 \mathrm{H}_6$ ?
(a) 18 moles
(b) 6 moles
(c) 3 moles
(d) 1 mole
Answer:
(b) 6 moles
Hint:
$\mathrm{C}_2 \mathrm{H}_6$ contains $6 \mathrm{H}$ atoms. 6 moles.
Question 69.
The molar mass of $\mathrm{Na}_2 \mathrm{SO}_4$ is
(a). 129
(b) 142
(c) 110
(d) 70
Answer:
(b) 142
Hint:
$\mathrm{Na}_2 \mathrm{SO}_4=$ Molar mass
$=(23 \times 2)+(32 \times 1)+(16 \times 4)$
$=46+32+64=142$
Question 70 .
Match the List-I with List-Il using the correct code given below the list.

Answer:
(b) 344
Question 71.
Ore mole of $\mathrm{CO}_2$ contains
(a) $6.02 \times 10^{23}$ atoms of $\mathrm{C}$
(b) $3 \mathrm{~g}$ of $\mathrm{CO}_2$
(c) $6.0210^{23}$ atoms of $\mathrm{O}$
(d) $18.1 \times 10^{23}$ molecules of $\mathrm{CO}_2$
Answer:
(a) $6.02 \times 10^{23}$ atoms of $\mathrm{C}$
Question 72 .
5.6 liters of oxygen at STP is equivalent to .........
(a) 1 mole
(b) $1 / 4$ mole
(c) $1 / 8$ mole
(d) $1 / 2$ mole
Answer:
(b) $1 / 4$ mole
Hint:
22.4 litres of $\mathrm{O}_2=1$ mole
$\therefore 5.6$ litres of $\mathrm{O}_2=\frac{1}{22.4} \times 5.6=0.25$ mole $=1 / 4$ mole.
Question 73.
How many grams are contained in 1 gram atom of $\mathrm{Na}$ ?
(a) $13 \mathrm{~g}$
(b) $1 \mathrm{~g}$
(c) $23 \mathrm{~g}$
(d) $1 / 23 \mathrm{~g}$
Answer:
(c) $23 \mathrm{~g}$
Hint:
1 gram atom of $\mathrm{Na}$

$\mathrm{Na}=$ Atomic mass $23 \mathrm{~g}$ (or) $23 \mathrm{amu}$
1 gram atom of $\mathrm{Na}=1 \mathrm{~mole}=23 \mathrm{~g}$.
Question 74.
$12 \mathrm{~g}$ of $\mathrm{Mg}$ will react completely with an acid to give
(a) 1 mole of $\mathrm{O}_2$
(b) $1 / 2$ mole of $\mathrm{H}_2$
(c) I mole of $\mathrm{H}_2$
(d) 2 mole of $\mathrm{H}_2$
Answer:
(b) $1 / 2$ mole of $\mathrm{H}_2$
Hint:

Question 75.
which of the following has the highest mass?
(a) I g atom of $\mathrm{C}$
(b) $1 / 2$ mole of $\mathrm{CH}_4$
(c) $10 \mathrm{ml}$ of water
(d) $3.011 \times 10^{23}$ atoms of oxygen
Answer:
(a) I g atom of C
(a) $1 \mathrm{~g}$ atom of $\mathrm{C}=12 \mathrm{~g}$
(b) $1 / 2$ mole of $\mathrm{CH}_4=\frac{12+4}{2}=8 \mathrm{~g}$.
(c) $10 \mathrm{ml}$ of water $\left(\mathrm{H}_2 \mathrm{O}\right)=1 \times 10=10 \mathrm{~g}$
(d) $3.011 \times 10^{23}$ atoms of oxygen $=0.5$ mole of oxygen $=8 \mathrm{~g}$
Question 76.
The empirical formula of sucrose is
(a) $\mathrm{CH}_2 \mathrm{O}$
(b) $\mathrm{CHO}$
(c) $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$
(d) $\mathrm{C}\left(\mathrm{H}_2 \mathrm{O}\right)_2$
Answer:
(a) $\mathrm{CH}_2 \mathrm{O}$
Hint:
Sucrose Molecular formula $=\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$
$
\text { E.F. }=\frac{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}{12}=\mathrm{CH}_2 \mathrm{O}
$
Question 77.
The number of grams of oxygen in $0.10 \mathrm{~mol}$ of $\mathrm{Na}_2 \mathrm{CO}_3 .10 \mathrm{H}_2 \mathrm{O}$ is
(a) $20.8 \mathrm{~g}$
(b) $18 \mathrm{~g}$
(c) $108 \mathrm{~g}$
(d) $13 \mathrm{~g}$
Answer:
(a) $20.8 \mathrm{~g}$

Hint:
$\mathrm{Na}_2 \mathrm{CO}_2 \cdot 10 \mathrm{H}_2 \mathrm{O}=1$ mole
1 mole of $\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}$ contains 13 oxygen atoms.
Mass of 13 oxygen atoms $=13 \times 16=208$
1 mole of $\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}$ contains $208 \mathrm{~g}$ of oxygen.
$\therefore 0.10$ mole of $\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}$ contains $\frac{208}{1} \times 0.12=08 \mathrm{~g}$.
Question 78.
The mass of an atom of nitrogen is
(a) $\frac{14}{6.023 \times 10^{23}} \mathrm{~g}$
(b) $\frac{28}{6.023 \times 10^{23}} g$
(c) $\frac{1}{6.023 \times 10^{23}} \mathrm{~g}$
(d) $14 \mathrm{amu}$
Answer:
(a) $\frac{14}{6.023 \times 10^{23}} g$
Mass of 1 atom $=\frac{\text { Atomic mass }}{\text { Avogadro number }}=\frac{14}{6.023 \times 10^{23}}$
Question 79.
Which of the following halogens do not exhibit positive oxidation number in its compounds?
(a) Fluorine
(b) Chlorine
(c) Iodine
(d) Bromine
Answer:
(a) Fluorine
Question 80 .
Which of the following is the most powerful oxidising agent?
(a) $\mathrm{KMnO}_4$
(b) $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
(c) $\mathrm{O}_3$
(d) $\mathrm{H}_2 \mathrm{O}_2$
Answer:
(a) $\mathrm{KMnO}_4$
Question 81 .
On the reaction $2 \mathrm{Ag}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Ag}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2$. Sulphuric acid acts as

(a) oxidising agent
(b) reducing agent
(c) a catalyst
(d) an acid as well as an oxidant
Answer:
(d) an acid as well as an oxidant
Question 82 .
The oxidation number of carboxylic carbon atom in $\mathrm{CH}_3 \mathrm{COOH}$ is
(a) +2
(b) +4
(c) +1
(d) +3
Answer:
(d) +3
$\mathrm{CH}_3 \mathrm{COOH}$
$-3+3+x-4+1$
$\mathrm{x}-3=0$
$x=+3$
Carboxylic carbon oxidation number $=+3$
Question 83.
When methane is burnt in oxygen to produce $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$, the oxidation number of carbon changes by
(a) -8
(b) +4
(c) zero
(d) +8
Answer:
(b) +4

Question 84.
The oxidation number of carbon is zero in
(a) $\mathrm{HCHO}$
(b) $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$
(c) $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
(d) all the above
Answer:
(d) all the above
Question 85 .
The oxidation number of $\mathrm{Fe}$ in $\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$ is ........
(a) +2
(b) +3
(c) $+2,+3$
(d) $\mathrm{O}$
Answer:
(b) +3
$\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$
$
\begin{aligned}
& 2 x-6=0 \\
& x=+3
\end{aligned}
$
Question 86.
Among the following molecules in which Chlorine shows maximum oxidation state?
(a) $\mathrm{Cl}_{21}$
(b) $\mathrm{KCl}$
(c) $\mathrm{KClO}_3$
(d) $\mathrm{Cl}_2 \mathrm{O}_7$
Answer:
(d) $\mathrm{Cl}_2 \mathrm{O}_7$
$
\begin{aligned}
& \mathrm{Cl}_2 \mathrm{O}_7 \\
& 2 \mathrm{x}-14=0 \\
& 2 \mathrm{x}=+14 \\
& \mathrm{x}=+7
\end{aligned}
$

Question 87.
The oxidation number of carbon in $\mathrm{CH}_3 \rightarrow \mathrm{CH}_2 \mathrm{OH}$ is
(a) +2
(b) -2
(e) $\mathrm{O}$
(d) +4
Answer:
(b) -2
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
$2 \mathrm{x}+6-2=0$
$2 \mathrm{x}+4=0$
$2 \mathrm{x}=-4$
$\mathrm{x}=-2$
2 - MarkQuestions
I. Write brief answer to the following questions:
Question 1.
State Avogadro's Hypothesis.
Answer:
It states that 'Equal volume of all gases under the same conditions of temperature and pressure contain the same number of molecules'.
Question 2.
What is molar volume?
Answer:
Molar volume is the volume occupied by one mole of a substance in the gaseous state at STP. It is equal to $2.24 \times 10^{-2} \mathrm{~m}^3(22.4 \mathrm{~L})$
Question 3.
The approximate production of $\mathrm{Na}_2 \mathrm{CO}_3$ per month is $424 \times 10^6 \mathrm{~g}$ while that of methyl alcohol is $320 \times 10^6$ g. Which is produced more in terms of moles?

Answer:
$
\mathrm{Na}_2 \mathrm{CO}_3 \text { mass }=424 \times 10^6 \mathrm{~g}
$
Molecular mass of $\mathrm{Na}_2 \mathrm{CO}_3=(23 \times 2)+12+(16 \times 3)$
$
\begin{aligned}
& =46+12+48 \\
& =106 \mathrm{~g}
\end{aligned}
$
No. of moles of $\mathrm{Na}_2 \mathrm{CO}_3$
$
=\frac{\text { Mass of } \mathrm{Na}_2 \mathrm{CO}_3}{\text { Molecular mass of } \mathrm{Na}_2 \mathrm{CO}_3}=\frac{424 \times 10^6 \mathrm{~g}}{106 \mathrm{~g}}
$
$=4 \times 10^6$ moles
Methyl alcohol mass $=320 \times 10^6 \mathrm{~g}$
Molecular mass of $\mathrm{CH}_3 \mathrm{OH}=12+(1 \times 4)+16=32 \mathrm{~g}$ $=12+4+16=32 \mathrm{~g}$
No. of moles of Methyl alcohol $=\frac{\text { Mass of Methyl alcohol }}{\text { Molecular mass of Methyl alcohol }}=\frac{320 \times 10^6 \mathrm{~g}}{32 \mathrm{~g}}$ $=10 \times 10^6$ moles.
$\therefore$ Methyl alcohol is more produced in terms of moles.
Question 4.
Calculate number of moles of carbon atoms in three moles of ethane.
Answer:
Ethane - Molecular formula $=\mathrm{C}_2 \mathrm{H}_6$
1 mole of ethane contains 2 atoms of carbon $\left(6.023 \times 10^{23} \mathrm{C}\right)$
$\therefore 3$ moles of ethane contains 6 atoms of Carbon.
$\therefore$ No. of moles of Carbon atoms $=3 \times 6.023 \times 10^{23}$ Carbon atoms.
$=18.069 \times 10^{23}$ Carbon atoms.
Question 5 .
Find the molecular mass of $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$
Answer:
Sum of Atomic mass of all elements = Molecular mass
Atomic mass of $\mathrm{Fe}=56.0$
Atomic mass of $\mathrm{S}=32.0$
Atomic mass of $4[\mathrm{O}]=64.0$
Atomic mass of $14[\mathrm{H}]=14.0$
Atomic mass of $7[\mathrm{O}]=112.0=278.0$
Molecular mass of $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}=278 \mathrm{~g}$.
Question 6.
Mass of one atom of an element is $6.66 \times 10^{23} \mathrm{~g}$. How many moles of element are there in $0.320 \mathrm{~kg}$ ?
Answer:
Mass of one atom of an element $=6.66 \times 10^{23} \mathrm{~g}$
No. of moles $=\frac{\text { Mass }}{\text { Molecularmass }} 3$
Molecular mass $=$ Mass of 1 atom $x$ Avogadro number
$
6.66 \times 10^{23} \times 6.023 \times 10^{23}
$

$
=6.66 \times 6.023=40.11318
$
Number of moles $=\frac{\text { Mass }}{\text { Molecularmass }}=\frac{0.320 \mathrm{~kg} \times 10^3}{40}=8 \mathrm{moles}$.
Question 7.
How many moles of glucose are present in $720 \mathrm{~g}$ of glucose?
Answer:
Glucose $=\mathrm{C}_6 \mathrm{H}_4 \mathrm{O}_4$
Molecular mass of Glucose $=(12 \times 6)+(1 \times 12)+(16 \times 6)$
$=72+12+96=180$
$\frac{720}{180}=4$ moles.
Question 8.
Calculate the weight of 0.2 mole of sodium carbonate.
Answer:
Sodium carbonate $=\mathrm{Na}_2 \mathrm{CO}_3$
Molecular mass of $\mathrm{Na}_2 \mathrm{CO}_3=(23 \times 2)+(12 \times 1)+(16 \times 3)$
$
=46+12+48=106 \mathrm{~g}
$
Mass of 1 mole of $\mathrm{Na}_2 \mathrm{CO}_3=\frac{106 \times 0.2}{1}=21.2 \mathrm{~g}$
Question 9.
What do you understand by the terms acidity and basicity?
Answer:
Acidity:
The number of hydroxyl ions present in one mole of a base is known as the acidity of the base.
Basicity:
The number of replaceable hydrogen atoms present in a molecule of the acid is referred to as its basicity.
Question 10.
Calculate the equivalent mass of bicarbonate ion.
Answer:
Bicarbonate ion $=\mathrm{HCO}_3$
Equivalent mass of an ion $=\frac{\text { Formula mass of the ion }}{\text { Negative charge of the ion }}$
Formula mass of $\mathrm{HCO}_3=1+12+48=61$

Question 11.
Calculate the equivalent mass of barium hydroxide.

Answer
Barium hydroxide $=\mathrm{Ba}(\mathrm{OH})_2$
Molecular mass of $\mathrm{Ba}(\mathrm{OH})_2=137+(16 \times 2)+(1 \times 2)$ $=171.0 \mathrm{~g} / \mathrm{mol}$
Acidity $=2$
Equivalent mass of $\mathrm{Ba}(\mathrm{OH})_2=$ $=\frac{171.0}{2}=85.5$
Question 12 .
Calculate the equivalent mass of hydrated sodium carbonate.
Answer:
Hydrated sodium carbonate $=\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}$
Molecular mass of $\mathrm{Na}_2 \mathrm{CO}_3 .10 \mathrm{H}_2 \mathrm{O}=(23 \times 2)+(2 \times 1)+(16 \times 13)+(1 \times 20)$
$
\begin{aligned}
& =46+12+208+20=286 \\
& \text { Equivalent mass of } \mathrm{Na}_2 \mathrm{CO}_3 .10 \mathrm{H}_2 \mathrm{O}=\frac{\text { Molecularmass }}{\text { Acidity }} \\
& =\frac{286}{2}=143
\end{aligned}
$
Question 13 .
What do you understand by the terms empirical formula and molecular formula?
Answer:
Empirical Formula:
- It is the simplest formula.
- It shows the ratio of number of atoms of different elements in one molecule of the compound.
Molecular Formula:
- It is the actual formula.
- It shows the actual number of different types of atoms present in one molecule of the compound.

Question 14.
Boric acid, $\mathrm{H}_3 \mathrm{BO}_3$ is a mild antiseptic and is often used as an eye wash. A sample contains $0.543 \mathrm{~mol}$ $\mathrm{H}_3 \mathrm{BO}_3$. What is the mass of boric acid in the sample?
Answer:
Molecular mass of $\mathrm{H}_3 \mathrm{BO}_3=(1 \times 3)+(11 \times 1)+(16 \times 3)=62$
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass $\mathrm{x}$ mole $=62 \times 0.543$ $=33.66 \mathrm{~g}$
Question 15.
A compound contains $50 \%$ of $\mathrm{X}$ (atomic mass 10) and $50 \% \mathrm{Y}$ (atomic mass 20). Give its molecular formula
Answer:

$\therefore$ The Empirical Formula is $\mathrm{X}_3 \mathrm{Y}$
Empirical Formula mass $=20+20=40$
Molecular mass $=$ Sum of atomic mass $=40$
$\mathrm{n}=1$, Molecular formula $=(\text { Empirical Formula })_{\mathrm{n}}=\left(\mathrm{X}_2 \mathrm{Y}\right)_1=\mathrm{X}_2 \mathrm{Y}$
Question 16.
Calculate the mass of sodium (in $\mathrm{kg}$ ) present in $95 \mathrm{~kg}$ of a crude sample of sodium nitrate whose percentage purity is $70 \%$.
Answer:
Sodium Nitrate $=\mathrm{NaNO}_3$
Molecular mass of Sodium Nitrate $=23+14+48=85$

$100 \%$ pure $85 \mathrm{~g}$ of $\mathrm{NaNO}_3$ contains $23 \mathrm{~g}$ of Sodium.
$100 \%$ pure $95 \times 103 \mathrm{~g}$ of $\mathrm{NaNO}_3$ will contains $\frac{23}{85} \times 95 \times 10^3$ $=25.70 \times 10^3 \mathrm{~g}$ of Sodium.
$100 \%$ pure $\mathrm{NaNO}_3$ contains $25.70 \times 10^3 \mathrm{~g}$ of Sodium.
$\therefore 70 \%$ pure $\mathrm{NaNO}_3$ will contains $=$ img
$=17990 \mathrm{~g}$ (or) $17.99 \mathrm{Kg}$ of $\mathrm{Na}$.
Question 17.
Define matter. What are the types of matter?
Answer:
- A matter is anything which has mass and occupies space.
- Matters exist in all three states such as solid, liquid and gas.
Question 18.
Prove that states of matter are inter convertible.
Answer:
States of matter are inter convertible by changing temperature and pressure.
Question 19.
What is meant by Plasma state? Give an example.
Answer:
Gaseous state of matter at very high temperature containing gaseous ions and free electron is referred to as the Plasma state. e.g. Lightning.
Question 20.
Differentiate an element and an atom.
Answer:
- An atom is the ultimate smallest electrically neutral, being made up of fundamental particles such as proton, neutron and electron.
- An element consists of only one type of atoms. Elements are further divided into metals, non-metals, and metaloids.
Question 21.
Distinguish between a molecule and a compound
Answer:
Molecule:
- A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence.
- e.g. $\mathrm{Na}$ - Mono atomic molecule $\mathrm{O}_2$ - Diatomic molecule $\mathrm{P}_4$ - Poly atomic molecule
Compound:

- A molecule which contains two or more atoms of different elements are called a compound molecule.
- e.g. $\mathrm{CO}_2$ - Carbon dioxide $\mathrm{CH}_4$ - Methane $\mathrm{H}_2 \mathrm{O}$ - Water
Question 22.
Chlorine has fractional average atomic mass. Justify this statement.
Answer:
Chlorine molecule has two isotopes as in ${ }_{17} \mathrm{Cl}^{35},{ }_{17} \mathrm{Cl}^{37}$ in the ratio of $77: 23$, so when we are calculating the average atomic mass, it becomes fractional.
The average relative atomic mass of Chlorine $=\frac{(35 x 77)+(37 x 23)}{100}=35.46 \mathrm{amu}$
Question 23.
Define molecular mass of a substance.
Answer:
Molecular mass of a substance (element or compound) represents the number of times the molecule of that substance is heavier than $1 / 12^{\text {th }}$ of the mass of an atom of $\mathrm{C}-12$ isotope. Molecular mass $=2 \times$ Vapour density
Question 24.
Calculate the molecular mass of Sulphuric acid $\left(\mathrm{H}_2 \mathrm{SO}_4\right)$. Element
Answer:

Question 25 .
Define Avogadro Number
Answer:
Arogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Arogadro number $(\mathrm{N})=6.023 \times 10^{23}$
Question 26.
Calculate the number of moles present in $60 \mathrm{~g}$ of ethane.
Answer:
Molar Mass of ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)=24+6=30$
Number of moles in $60 \mathrm{~g}$ of ethane $=\frac{60}{30}=2$ moles.
Question 27.
Calculate the equivalent mass of Copper. (Atomic mass of copper $=63.5$ )
Answer:
Equivalent mass $=\frac{\text { Atomicmass }}{\text { Valency }}$
Equivalent mass of Copper $=\frac{63.5}{2}=31.75 \mathrm{~g} \mathrm{eq}^{-1}$
Question 28.
Calculate the equivalent mass of (i) Sulphate ion (ii) Phosphate ion.
Answer:

(i) Sulphate ion $\left(\mathrm{SO}_4^{2-}\right)$.
$
\text { Equivalent mass of Sulphate }=\frac{\text { Molar mass of Sulphate ion }}{\text { Charge }}=\frac{32+64}{2}=\frac{96}{2}=48 \mathrm{~g} \mathrm{eq}^{-1}
$

(ii) Phosphate ion $\left(\mathrm{PO}_4{ }^{3-}\right)$
$
\text { Molar mass of Phosphate ion }=\frac{\text { Molar mass of Phosphate ion }}{\text { Charge }}=\frac{31+64}{3}=\frac{95}{2}=31.6=31.6 \mathrm{~g} \mathrm{eq}^{-1}
$

Question 29.
Calculate the equivalent mass of sulphuric acid.
Answer:
Sulphuric acid $=\mathrm{H}_2 \mathrm{SO}_4$
Molar mass of Sulphuric acid $=2+32+64=96$
Basicity of Sulphuric acid $=2$

Question 30.
How many moles of hydrogen is required to produce 20 moles of ammonia?
Answer:
$
3 \mathrm{H}_2+\mathrm{N}_2 \rightarrow 2 \mathrm{NH}_3
$
A per stoichiometric equation,
No. of moles of hydrogen required for 2 moles of ammonia 3 moles
No. of moles of hydrogen required for 20 moles of ammonia $=\frac{3}{2} \times 20=30$ moles.
Question 31 .
Calculate the amount of water produced by the combustion of $32 \mathrm{~g}$ of methane.
Answer:

As per stoichiometric equation,
$16 \mathrm{~g}$ of methane produces $36 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$
$\therefore 32 \mathrm{~g}$ of methane will produce $=\frac{36}{16} \times 32=72 \mathrm{~g}$ of water.
Question 32.
How much volume of Carbon dioxide is produced when $25 \mathrm{~g}$ of calcium carbonate is heated completely under standard conditions?
Answer:

$
\begin{aligned}
& 100 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \text { produces } 22.4 \mathrm{~L} \text { of } \mathrm{CO}_2 . \\
& \therefore 25 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \text { will produce }=\frac{22.4}{100} \times 25=5.6 \mathrm{~L}_{\text {of }} \mathrm{CO}_2 \text {. } \\
&
\end{aligned}
$
Question 33.
How much volume of chlorine is required to prepare $89.6 \mathrm{~L}$ of $\mathrm{HCl}$ gas at STP?
Answer:

$2 \times 22.4 \mathrm{~L}$ of $\mathrm{HCl}$ is produced by $22.4 \mathrm{~L}_{\text {of }} \mathrm{Cl}_2$.
$\therefore 89.6 \mathrm{~L}$ of $\mathrm{HCl}$ will be produced by $=$ img $=89.6 \mathrm{~L}=44.8 \mathrm{~L}$ of chlorine
Question 34 .
What is meant by limiting reagent?
Answer:
A large excess of one reactant is supplied to ensure the more expensive reactant is completely converted to the desired product. The reactant used up first in a reaction is called the limiting reagent.
Question 35.
On the formation of $\mathrm{SF}_6$ by the direct combination of $\mathrm{S}$ and $\mathrm{F}_2$, which is the limiting reagent? Prove it.
Answer:
$\mathrm{SF}_6$ is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of $\mathrm{S}$ is allowed to react with 12 moles of Fluorine.
$
\mathrm{S}_{(1)}+3 \mathrm{~F}_{2(\mathrm{~g})} \rightarrow \mathrm{SF}_{(\mathrm{g})}
$
As per the stoichiometric reaction, one mole of $\mathrm{S}$ reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.
$\therefore$ It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.
Question 36.
Mention any 4 redox reaction that takes place in our daily life.
Answer:
1. Burning of cooking gas, wood
2. Rusting of iron articles
3. Electroplating
4. Galvanic and electrolytic cells
Question 37.
Calculate the oxidation number of underlined elements in the following.
1. $\mathrm{KMnO}_4$
2. $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$
Answer:
1. $\mathrm{KMnO}_4$
$
\begin{aligned}
& 1(+1)+\mathrm{x}+4(-2)=0 \\
& \mathrm{x}-7=0 \\
& \therefore \mathrm{x}=+7
\end{aligned}
$
Oxidation state of $\mathrm{Mn}=+7$.
2. $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$
$
\begin{aligned}
& 2 x+7(-2)=-2 \\
& 2 x-14=-2 \\
& 2 x=+12
\end{aligned}
$

$
\therefore \mathrm{x}=+6
$
Oxidation state of $\mathrm{Cr}=+6$.
Question 38.
If 10 volumes of $\mathrm{H}_2$ gas react with 5 volumes of $\mathrm{O}_2$ gas, how many volumes of water vapour would be produced?
Answer:
Thus 2 volumes of $\mathrm{H}_2$ reacts with 1 volume of $\mathrm{O}_2$ to produce 2 volumes of $\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$.
10 volumes of $\mathrm{H}_2$ would react with 5 volumes of $\mathrm{O}_2$ to produce 10 volumes of $\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$
Thus 10 volumes of $\mathrm{H}_2 \mathrm{O}$ will be produced.
Question 39.
Which one of the following will have largest number of atoms?
1. $1 \mathrm{~g}$ of $\mathrm{Au}(\mathrm{s})$
2. $1 \mathrm{~g}$ of $\mathrm{Na}(\mathrm{s})$
3. $1 \mathrm{~g}$ of $\mathrm{Li}(\mathrm{s})$
4. $1 \mathrm{~g}$ of $\mathrm{Cl}_2(\mathrm{~g})$
Answer:
1. Molar mass of $\mathrm{Au}=197 \mathrm{~g} \mathrm{~mol}^{-1}$.
No: of atoms in $1 \mathrm{~g}$ of $\mathrm{Au}=\frac{1}{197} \times 6.023 \times 10^{23}$
2. Molar mass of $\mathrm{Na}=23 \mathrm{~g} \mathrm{~mol}^{-1}$.
No of atoms in $1 \mathrm{~g}$ of $\mathrm{Na}=\frac{1}{23} \times 6.023 \times 10^{23}$
3. Molar mass of $\mathrm{Li}=7 \mathrm{~g} \mathrm{~mol}^{-1}$
No. of atoms in $1 \mathrm{~g}$ of $\mathrm{Li}=\frac{1}{7} \times 6.023 \times 10^{23}$
4. Molar mass of $\mathrm{Cl}_2=35.5 \mathrm{~g} \mathrm{~mol}^{-1}$
No. of atoms in $1 \mathrm{~g}$ of $\mathrm{Cl}_2=\frac{1}{35.5} \times 6.023 \times 10^{23}$
Comparing the number of atoms, the largest number of atoms will be present in $1 \mathrm{~g}$ of $\mathrm{Li}$. Since the mass is same in each case, the element with the lowest molar mass would have the largest number of atoms.
$\therefore \mathrm{Li}$ with lowest molar mass would have the largest number of atoms.
Question 40 .
What will be the mass of one ${ }^{12} \mathrm{C}$ atom in $\mathrm{g}$ ?
Answer:
Molar mass of ${ }^{12} \mathrm{C}=12.00 \mathrm{~g} \mathrm{~mol}^{-1}$.
$\therefore$ Mass of $6.023 \times 10^{23}$ carbon atom $=12.0 \mathrm{~g}$

$
\therefore \text { Mass of } 1 \text { carbon atom }=\frac{12}{6.023 \times 10^{23}}=1.992 \times 10 \mathrm{~g} \text {. }
$
Question 41.
Justify the following reaction is a redox reaction.
Answer:
$
\begin{aligned}
& \mathrm{CuO}_{(\mathrm{s})}+\mathrm{H}_{2_{(\mathrm{g})}} \rightarrow \mathrm{Cu}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \\
& (+2)(-2) \quad(0) \quad(0) \quad(+1)(-2) \\
& \mathrm{CuO}_{(\mathrm{s})}+\mathrm{H}_{2_{(\mathrm{g})}} \rightarrow \mathrm{Cu}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \text {. } \\
&
\end{aligned}
$
In the above reaction, oxygen is removed from $\mathrm{CuO}$. So $\mathrm{CuO}$ gets reduced. Oxygen is added to $\mathrm{H}_2$ to form water. $\mathrm{So}_2 \mathrm{H}_2$ gets oxidised. i.e. In $\mathrm{CuO}$, oxidation number of $\mathrm{Cu}+2$ is reduced to $\mathrm{O}$ whereas in $\mathrm{H}_2$, oxidation of $\mathrm{H}_2 \mathrm{O}$ is increased to +1 . So the above reaction is a redox reaction.
3 - MarkQuestions
Question 1.
Distinguish among the different physical states of matter.
Answer:
Differences among three physical states of matter (solid, liquid and gas) are as follows

Question 2.
Define equivalent mass of a salt.
Answer:
Equivalent mass of a salt:
It is defined as the number of parts by mass of the salt that is produced by the neutralization of one equivalent of an acid by a base. Therefore the equivalent mass of the salt is equal to its molar mass.
Question 3.
How much copper can be obtained from $100 \mathrm{~g}$ of anhydrous copper sulphate?
Answer:
Anhydrous copper sulphate $=\mathrm{CuSO}_4$
Molecular mass of $\mathrm{CuSO}_4=63.5+32+(16 \times 4)$
$=63.5+32+64$
$=159.5 \mathrm{~g}$
$159.5 \mathrm{~g}$ of $\mathrm{CuSO}_4$ contains $63.5 \mathrm{~g}$ of copper.
$\therefore 100 \mathrm{~g}$ of $\mathrm{CuSO}_4$ contains $\frac{6.35}{159.5} \times 100=0.39811 \times 100=39.81 \mathrm{~g}$ of Copper.
Question 4.
Calculate the equivalent mass of hydrated ferrous sulphate.
Answer:
Hydrated ferrous sulphate $=\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$
Ferrous sulphate - Reducing agent
Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation.
$2 \mathrm{FeSO}_4+\mathrm{H}_2 \mathrm{SO}_4+[\mathrm{O}] \rightarrow \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}$
$2 \times 152 \mathrm{~g} \quad 16 \mathrm{~g}$
16 parts by mass of oxygen oxidised $304 \mathrm{~g}$ of $\mathrm{FeSO}_4$.
8 parts by mass of oxygen will oxidise $\frac{304}{16} \times 8$ parts by mass of $\mathrm{FeSO}_4$.
$=152$
Equivalent mass of Ferrous sulphate (Anhydrous) $=152$
Equivalent mass of crystalline Ferrous sulphate $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}=152+126=278$

Question 5.
Give difference between empirical and molecular formula
Answer:
Empirical Formula:
- Empirical formula is the simplest formula.
- It shows the ratio of number of atoms of different elements in one molecule of the compound.
- It is calculated from the percentage of composition of the various elements in one molecule.
- For example, Empirical formula of Benzene $=\mathrm{CH}$.
Molecular Formula:
- Molecular Formula is the actual formula.
- It shows the actual number of different types of atoms present in one molecule of the compound.
- It is calculated from the Empirical formula Molecular Formula = (Empirical formula $)_n$
- Molecular formula of Benzene $=\mathrm{C}_6 \mathrm{H}_6$.
Question 6.
A sample of hydrated copper sulphate is heated to drive off the water of crystallization, cooled and reweighed $0.869 \mathrm{~g}$ of $\mathrm{CuSO} 4 \mathrm{aH}_2 \mathrm{O}$ gave a residue of $0.556 \mathrm{~g}$. Find the molecular formula of hydrated copper sulphate.
Answer:
$0.869 \mathrm{~g}$ of $\mathrm{CuSO}_4 \cdot \mathrm{aH}_2 \mathrm{O}$ gave a residue of $0.556 \mathrm{~g}$ of Anhydrous $\mathrm{CuSO}_4$.
$\therefore$ Weight of a $\mathrm{H}_2 \mathrm{O}$ molecule $=0.869-0.556=0.313 \mathrm{~g}$
Molecular weight of $\mathrm{H}_2 \mathrm{O}=(1 \times 2)+16=2+16=18$
No. of moles of water $=\frac{\text { Mass }}{\text { Molecularmass }}$
$\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}-$ Molecular mass $=63.5+32+64+90=249.5 \mathrm{~g}$
$249.5 \mathrm{~g}$ of $\mathrm{CuSO}_4 .5 \mathrm{H}_2 \mathrm{O}$ on heating gives $159.5 \mathrm{~g}$ of $\mathrm{CuSO}_4$.
$0.869 \mathrm{~g}$ of $\mathrm{CuSO}_4 \cdot \mathrm{aH}_2 \mathrm{O}$ on heating gives $=\frac{159.5}{249.5} \times 0.869$
$=0.556 \mathrm{~g}$ of anhydrous $\mathrm{CuSO}_4 \therefore \mathrm{a}=5$
The molecular formula of hydrated copper sulphate $=\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$
Question 7.
Balance by oxidation number method: $\mathrm{Mg}+\mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
Step -1

Step - 2
$
\mathrm{Mg}+2 \mathrm{HNO}_3-\gg \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}
$
Step -3
To balance the number of oxygen atoms and hydrogen atoms $2 \mathrm{HNO}_3$ is multiplied by 2 .
$
\mathrm{Mg}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}
$
Step -4
To balance the number of hydrogen atoms, the $\mathrm{H}_2 \mathrm{O}$ molecule is multiplied by 2 .
$
\mathrm{Mg}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}
$
Question 8.
Explain about the classification of matter.
Answer:

Question 9.
Calculate the mass of the following atoms in amu,
(a) Helium (mass of $\mathrm{He}=6.641 \times 10^{-24} \mathrm{~g}$ )
(b) Silver (mass of $\mathrm{Ag}=1.790 \times 10^{-22} \mathrm{~g}$ )
$1 \mathrm{amu}=1.66056 \times 10^{-24}$
Answer:
(a) The mass of Helium atom in $\mathrm{amu}=\frac{6.641 \times 10^{-24}}{1.66056 \times 10^{-24}}=3.9992 \mathrm{amu}$.
(b) The mass of Silver atom in $\mathrm{amu}=\frac{1.790 \times 10^{-22}}{1.66056 \times 10^{-24}}=107.79 \mathrm{amu}$.
Question 10 .
Calculate the number of atoms present in $1 \mathrm{Kg}$ of gold.
Answer:
The atomic mass of Gold $=197 \mathrm{~g} \mathrm{~mol}^{-1}$.
$197 \mathrm{~g}$ of gold contains $6.023 \times 10^{23}$ atoms of gold.
$\therefore 1000$ g of gold will contain $=\frac{1000 \times 6.023 \times 10^{23}}{197}$
$=3.055 \times 10^{24}$ atoms of Gold.
Question 11.
Calculate the molar volume of $146 \mathrm{~g}$ of $\mathrm{HCl}$ gas and the number of molecules present in it.
Answer:
Molar mass of $\mathrm{HCl}=36.5 \mathrm{~g}$
The molar volume of $36.5 \mathrm{~g}$ (1 mole) of $\mathrm{HCl}=2.24 \times 10^2 \mathrm{~m}^3$.
$\therefore$ The volume of $146 \mathrm{~g}$ (4 moles) of $\mathrm{HCl}=\frac{2.24 \times 10^{-2}}{36.5} \times 146$
$=8.96 \times 10 \mathrm{~m}^3$
No. of molecules in $146 \mathrm{~g}$ of $\mathrm{HCl}=4 \mathrm{~N}$
$=4 \times$ Avogadro Number
$=4 \times 6.023 \times 10^{23}$
$=24.092 \times 10^{23}$
$=2.4092 \times 10^{24}$ molecules.

Question 12.
Calculate the molar mass of $20 \mathrm{~L}$ of gas weighing $23.2 \mathrm{~g}$ at STP.

Answer:
Weight of the substance $\times$ Molar volume
Molar mass $=$ Volume of the substance at STP
Molar volume at $\mathrm{STP}=2.24 \times 10^{-2} \mathrm{~m}^3=22.4 \mathrm{~L}$ (or) $22400 \mathrm{cc}$.
Molar mass of the gas at STP $=\frac{23.2 x 22.4}{20}=25.984 \mathrm{~g}$.
Question 13.
$0.6 \mathrm{~g}$ of a metal gives on oxidation $1 \mathrm{~g}$ of its oxide. Calculate its equivalent mass.
Answer:
Mass of metal $=0.6$
Mass of metal oxide $=1 \mathrm{~g}$
Mass of oxygen $=1-0.6=0.4 \mathrm{~g}$
$0.4 \mathrm{~g}$ of oxygen combines with $0.6 \mathrm{~g}$ of metal.
$\therefore 8 \mathrm{~g}$ of oxygen will combine with $=\frac{0.6}{0.4} \times 8$
Equivalent mass of the metal $=12 \mathrm{~g} \mathrm{eq}^{-1}$.
Question 14.
How would you calculate the equivalent mass of anhydrous oxalic acid and hydrated oxalic acid.
Answer:
In acid medium,

$16 \mathrm{~g}$ of oxygen is used for oxidation of $90 \mathrm{~g}$ of oxalic acid.
$
\therefore 8 \text { g of oxygen will oxidize }=\frac{90}{16} \mathrm{x} 8=45 \mathrm{~g} \mathrm{eq}^{-1} \text {. }
$
Equivalent mass of Anhydrous oxalic acid $=45 \mathrm{~g} \mathrm{eq}^{-1}$

$
=\frac{126}{2}=63 \mathrm{~g} \mathrm{eq}^{-1} .
$
Question 15 .
A compound on decomposition in the laboratory produces $24.5 \mathrm{~g}$ of nitrogen and $70 \mathrm{~g}$ of oxygen. Calculate the empirical formula of the compound.
Answer:

the empirical formula is $\mathrm{N}_2 \mathrm{O}_5$
Question 16.
What is the steps involve in the calculation of molecular formula from empirical formula?
Answer:
Molecular mass and empirical formula are used to deduce molecular formula of the compound.
Steps to calculate molecular formula:
- if Empirical formula is found out from the percentage composition of elements
- Empirical formula mass can be found from the empirical formula
- Molecular mass is found out from the given data
- Molecular formula $=(\text { Empirical formula })_{\mathrm{n}}$
Molecular mass
- where, $\mathrm{n}=$ Empirical formula mass
Question 17.
What is combination reaction? Give example.
When two or more substances combine to form a single substance, the reactions are called combination reactions.
$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$
Example:
$2 \mathrm{Mg}+\mathrm{O}_2 \rightarrow 2 \mathrm{MgO}$
Question 18.
What is decomposition reaction? Give two examples.
Answer:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
$\mathrm{AB} \rightarrow \mathrm{A}+\mathrm{B}$
Example:
$2 \mathrm{KCO}_3 \rightarrow 2 \mathrm{KCl}+\mathrm{O}_2$
$\mathrm{PCl}_5 \rightarrow \mathrm{PCl}_3+\mathrm{Cl}_2$

$\mathrm{AB} \rightarrow \mathrm{A}+\mathrm{B}$
Example:
$
\begin{aligned}
& 2 \mathrm{KCO}_3 \rightarrow 2 \mathrm{KCl}+\mathrm{O}_2 \\
& \mathrm{PCl}_5 \rightarrow \mathrm{PCl}_3+\mathrm{Cl}_2
\end{aligned}
$
Question 19.
What is displacement reactions? Give its types. Explain with example.
Answer:
The reactions in which one ion or atom in a compound is replaced (or substituted) by an ion or atom of the other element are called displacement reactions.
$
\mathrm{AB}+\mathrm{C} \rightarrow \mathrm{AC}+\mathrm{B}
$
Example:
Metal displacement
$
\mathrm{CuSO}_4+\mathrm{Zn} \rightarrow \mathrm{ZnSO}_4+\mathrm{Cu}
$
Example:
Non-metal displacement
$
2 \mathrm{KBr}+\mathrm{Cl}_2 \rightarrow 2 \mathrm{KCl}+\mathrm{Br}_2
$
Question 20 .
What is disproportionation reactions? Give example.
Answer:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
$
\begin{aligned}
& \mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3+3 \mathrm{NaH}_2 \mathrm{PO}_2 \\
& 2 \mathrm{HCHO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{OH}+\mathrm{HCOOH}
\end{aligned}
$
Question 21.
What are competitive electron transfer reaction? Give example.
Answer:
These are the reactions in which redox reactions take place in different vessels and it is andirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
$\mathrm{Zn}$ releases electrons to $\mathrm{Cu}$ and $\mathrm{Cu}$ releases electrons to Silver and so on.
$
\begin{aligned}
& \mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}{ }_{(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})} \text { (Here } \mathrm{Zn} \text {-oxidised; } \mathrm{Cu}^{2+}-\text { reduced) } \\
& \mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{2+} \text { (aq) }+2 \mathrm{Ag}_{(\mathrm{s})} \text { (Here } \mathrm{Cu} \text {-oxidised; } \mathrm{Ag}^{+}-\text {reduced) }
\end{aligned}
$
Question 22.
Balance the following equation using oxidation number method.
Answer:
$
\text { 1. } \mathrm{S}+\mathrm{HNO}_3 \rightarrow \mathrm{H}_2 \mathrm{SO}_4+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}
$

2. $\mathrm{S}+6 \mathrm{HNO}_3 \rightarrow \mathrm{H}_2 \mathrm{SO}_4+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
3. Balance the equation (except $\mathrm{O}$ and $\mathrm{H}$ ) $\mathrm{S}+6 \mathrm{HNO}_3 \rightarrow \mathrm{H}_2 \mathrm{SO}_4+6 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
4. Balance $\mathrm{O}$ atoms by adding $2 \mathrm{H}_2 \mathrm{O}$ $5+6 \mathrm{HNO}_3 \rightarrow \mathrm{H}_2 \mathrm{SO}_4+6 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}$
Question 23.
Determine the empirical formula of an oxide of iron which has $69.9 \%$ iron and $30.1 \%$ oxygen by mass.

Answer:

the empirical formula is $\mathrm{Fe}_2 \mathrm{O}_3$
Question 24.
In three moles of ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)$ calculate the following:
1. Number of moles of carbon atoms.
2. Number of moles of hydrogen atoms.
3. Number of molecules of ethane.
Answer:
(1) 1 mole of $\mathrm{C}_2 \mathrm{H}_6$ contains 2 moles of Carbon atoms.
$\therefore 3$ moles of $\mathrm{C}_2 \mathrm{H}_6$ will have 6 moles of Carbon atoms.
(2) 1 mole of $\mathrm{C}_2 \mathrm{H}_6$ contains 6 moles of Hydrogen atoms.
$\therefore 3$ moles of $\mathrm{C}_2 \mathrm{H}_6$ will have 18 moles of Hydrogen atoms.
(3) 1 mole of $\mathrm{C}_2 \mathrm{H}_6$ contains $6.023 \times 10^{23}$ number of molecules.
$\therefore 3$ moles of $\mathrm{C}_2 \mathrm{H}_6$ will contain $3 \times 6.023 \times 10^{23}$ molecules.
Question 25 .
Chlorine is prepared in the laboratory by treating manganese dioxide $\left(\mathrm{MnO}_2\right)$ with aqueous hydrochloric acid according to the reaction.
$
4 \mathrm{HCl}_{(\mathrm{aq})}+\mathrm{MnO}_{2_{(\mathrm{s})}} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(1)}+\mathrm{MnCl}_{2(\mathrm{aq})}+\mathrm{Cl}_{2(\mathrm{aq})}
$
How many grams of $\mathrm{HCl}$ react with 5.0 g of manganese dioxide? (Atomic mass of $\mathrm{Mn}=55 \mathrm{~g}$ ).
Answer:
1 mole of $\mathrm{MnO}_2=55+32=87 \mathrm{~g}$.
$87 \mathrm{~g}$ of $\mathrm{MnO}_2$ reacts with 4 moles of $\mathrm{HCl}$. i.e. $=4 \mathrm{x} 36.5=146 \mathrm{~g}$ of $\mathrm{HCl}$.
$\therefore 5 \mathrm{~g}$ of $\mathrm{MnO}_2$ will react with $\frac{146}{87} \times 5.0=8.40 \mathrm{~g}$.

Question 26.
The density of water at room temperature is $1.0 \mathrm{~g} / \mathrm{ml}$. How many molecules are there in a drop of water if its volume is $0.05 \mathrm{ml}$ ?
Answer:
Volume of drop of water $=0.05 \mathrm{ml}$
Mass of a drop of water $=$ Volume $\mathrm{x}$ Density
$=0.05 \mathrm{ml} \mathrm{x} 1.0 \mathrm{~g} / \mathrm{ml}$
$=0.05 \mathrm{~g}$
Molar mass of water $\left(\mathrm{H}_2 \mathrm{O}\right)=18 \mathrm{~g}$
$18 \mathrm{~g}$ of water $=1$ mole
$0.05 \mathrm{~g}$ of water $=\frac{1}{18} \times 0.05=0.0028 \mathrm{~mol}$.
No. of molecules present in one mole of water $=6.023 \times 10^{23}$
No. of molecules present in 0.0028 mole of water $=\frac{6.023 \times 10^{23} \times 0.0028}{1}=1.68 \times 10^{21}$ water molecules
Question 27.
Balance the following equation by oxidation number method. $\mathrm{MnO}_4^{-}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}$ (Acidic medium)
Answer:

to balance $\mathrm{O}$ and $\mathrm{H}$ atoms $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}^{+}$are added.
$
\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}
$
5-Mark Questions
I. Answer the following questions in detail

Question 1.
Define the following (a) equivalent mass of an acid (b) equivalent mass of a base (c) equivalent mass of an oxidising agent $(\mathrm{cl})$ equivalent mass of a reducing agent.
Answer:
(a) Equivalent mass of an acid:
Equivalent mass of an acid is the number of parts by mass of the acid which contains 1.008 part by mass of replaceable hydrogen atom.
Equivalent mass of an acid $=\frac{\text { Molar mass of the acid }}{\text { Basicity of the acid }}$
(b) Equivalent mass of a base:
It is defined as the number of parts by mass of the base which contains one replaceable hydroxyl ion or which completely neutralizes one gram equivalent of an acid.
Equivalent mass of a base $=\frac{\text { Molar mass of the base }}{\text { Acidity of the base }}$
(c) Equivalent mass of an oxidising agent:
It is defined as the number of parts by mass of an oxidising agent which can furnish 8 parts by mass of oxygen for oxidation.
(d) Equivalent mass of a reducing agent:
It is defined as the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent.
Question 2.
Calculate the percentage composition of the elements present in lead nitrate. How many $\mathrm{Kg}$ of 02 can be obtained from $50 \mathrm{~kg}$ of $70 \%$ pure lead nitrate?

Answer:
Lead nitrate $=\mathrm{Pb}\left(\mathrm{NO}_3\right)_2$
Molecular mass of lead nitrate $=207+(14 \times 2)+(16 \times 6)$ $=207+28+96=331 \mathrm{~g} / \mathrm{mol}$.
$331 \mathrm{~g}$ of lead nitrate contains $96 \mathrm{~g}$ of oxygen.
$\therefore 50 \times 10^3 \mathrm{~g}$ of lead nitrate will contain $\frac{96}{331} \times 50 \times 10^3$
$=14501.5 \mathrm{~g}$
$=14.501 \mathrm{Kg}$ of oxygen.
$100 \%$ pure lead nitrate contains $14.501 \mathrm{Kg}$ of oxygen.
$70 \%$ pure lead nitrate will contain $=\frac{14.501}{100} \times 70=10.15 \mathrm{Kg}$ of oxygen.
$\therefore 70 \%$ pure lead nitrate will contain $10.15 \mathrm{Kg}$ of oxygen.
Question 3.
Determine the empirical formula of a compound containing $\mathrm{K}=24.15 \%, \mathrm{Mn}=34.77 \%$ and rest is oxygen.

Answer:

empirical formula of a compound $=\mathrm{KMnO}_4$
Question 4.
Write the steps to be followed for writing empirical formula.
Answer:
Empirical formula shows the ratio of number of atoms of different elements in one molecule of the compound.
Steps for finding the Empirical formula:
The percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.
1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound.
2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements.
3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio.
4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.
5. Percentage of Oxygen $=100-$ Sum of the percentage masses of all the given elements.
Question 5.
An organic compound was found to contain carbon $=40.65 \%$, hydrogen $=8.55 \%$ and Nitrogen $=23.7 \%$. Its vapour density was found to be 29.5. What is the molecular formula of the compound?
Answer:

Empirical formula $=\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}$
Molecular mass $=2 \times$ vapour density $=2 \times 29.5=59.0$
Empirical formula mass of $\mathrm{C}_4 \mathrm{H}_5 \mathrm{NO}=24+5+14+16=59$
$\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{59}{59}=1$
Empirical formula mass 59
$\therefore$ Molecular formula $=(\text { Empirical formula })_n$
$=\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}\right)_1$
Molecular formula $=\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}$
Question 6.
Calculate the empirical and molecular formula of a compound containing $32 \%$ carbon, $4 \%$ hydrogen and rest oxygen. Its vapour density is 75 .
Answer:

Empirical formula $=\mathrm{C}_2 \mathrm{H}_3 \mathrm{O}_3$
Empirical formula mass $=24+3+48=75$
Molecular mass $=2 \times$ Vapour density $=2 \times 75=150$

$\begin{aligned}
& \mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{150}{75}=2 \\
& \text { Molecular formula }=(\text { Empirical formula })_{\mathrm{n}}
\end{aligned}$

Molecular formula $=\mathrm{C}_2 \mathrm{H}_3 \mathrm{O}_3 \times 2$
Molecular formula $=\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}_6$
Question 7.
Explain the different types of redox reactions with example.
Answer:
Redox reactions are classified into the following types:
(1) Combination reactions:
When two or more substances combine to form a single substance, the reactions are called combination reactions.
Example:
$
2 \mathrm{Mg}+\mathrm{O}_2 \rightarrow 2 \mathrm{MgO}
$
(2) Decomposition reactions:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
Example:
$
2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2
$
(3) Displacement reactions:
The reactions in which one ion or atom in a compound is replaced by an ion or atom of the other element are called displacement reactions.
Example:
$
\mathrm{CuSO}_4+\mathrm{Zn} \rightarrow \mathrm{ZnSO}_4+\mathrm{Cu}
$
(4) Disproportionation reactions:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
$
2 \mathrm{HCHO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{OH}+\mathrm{HCOOH}
$
(5) Competitive Electron transfer reactions:
These are the reactions in which redox reactions take place in different vessels and it is andirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
$
\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}{ }_{(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})} \text { Here } \mathrm{Zn} \text {-oxidised; } \mathrm{Cu}^{2+} \text { reduced }
$
Question 8.
Write the steps to be followed while balancing redox equation by oxidation number method.
Answer:
Oxidation number method:
This method is based on the fact that
Number of electrons lost by atoms $=$ Number of electrons gained by atoms
Steps to be followed while balancing Redox reactions by Oxidation Number method:

1. Write skeleton equation representing redox reaction
2. Write the oxidation number of atoms undergoing oxidation and reduction.
3. Calculate the increase or decrease in oxidation numbers per atom.
4. Make increase in oxidation number equal to decrease in oxidation number by multiplying the formula of oxidant and reductant by suitable numbers.
5. Balance the equation atomically on both sides except $\mathrm{O}$ and $\mathrm{H}$ atoms.
6. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
7. Add required number of $\mathrm{H}+$ ions to the side deficient in hydrogen atom if the reaction is in acidic medium.
8. For reactions in basic medium, add $\mathrm{H}_2 \mathrm{O}$ molecules to the side deficient in hydrogen atoms and simultaneously add equal number of $\mathrm{OH}$ ions on the other side of the equation.
9. Finally balance the equation by cancelling common species present on both sides of the equation.
Question 9.
Balance the following equation by oxidation number method:
Answer:
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{KCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{KHSO}_4+\mathrm{CrO}_2 \mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O}$
Step - 1

$
\begin{aligned}
& \text { Step }-2 \\
& \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{KCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{KHSO}_4+2 \mathrm{CrO}_2 \mathrm{C}_2+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Step 3.
To balance $\mathrm{Cl}$ atom, $\mathrm{KCl}$ is multiplied by 4
$
\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{KCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{CrO}_2 \mathrm{Cl}_2+\mathrm{KHSO}_2+\mathrm{H}_2 \mathrm{O}
$
Step 4.
To balance $\mathrm{K}$ atom, $\mathrm{KHSO}_4$ is multiplied by 6 .
$
\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O}_7+4 \mathrm{KCl}+\mathrm{H} 2 \mathrm{SO}_4 \rightarrow 2 \mathrm{CrO}_2 \mathrm{Cl}_2+6 \mathrm{KHSO}_4+\mathrm{H}_2 \mathrm{O}
$
Step 5.
To balance $\mathrm{O}$ and $\mathrm{H}$ atoms, $\mathrm{HSO}_4$ is multiplied by $6, \mathrm{H} 20$ is multiplied by 3 .

$
\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{KCl}+6 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{CrO}_2 \mathrm{Cl}_2+6 \mathrm{KHSO}_4+3 \mathrm{H}_2 \mathrm{O}
$
(2) $\mathrm{P}+\mathrm{HNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
Step -1

Step 2.
$
\mathrm{P}+5 \mathrm{HNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4+5 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}
$
(3) $\mathrm{CuO}+\mathrm{NH}_3 \rightarrow \mathrm{Cu}+\mathrm{N}_2+\mathrm{H}_2 \mathrm{O}$
Step 1.

Step 2.
$
3 \mathrm{CuO}+2 \mathrm{NH}_3 \rightarrow 3 \mathrm{Cu}+\mathrm{N}_2+\mathrm{H}_2 \mathrm{O}
$
Step 3.
To balance $\mathrm{O}$ and $\mathrm{N}$, water is multiplied by 3 .
$
3 \mathrm{CuO}+2 \mathrm{NH}_3 \rightarrow 3 \mathrm{Cu}+\mathrm{N}_2+3 \mathrm{H}_2 \mathrm{O}
$
(4) $\mathrm{Zn}+\mathrm{HNO}_3 \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{NH}_4 \mathrm{NO}_3+\mathrm{H}_2 \mathrm{O}$ Step-1

Step 2.
$
4 \mathrm{Zn}+\mathrm{HNO}_3 \rightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_3+\mathrm{NH}_4 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}
$
Step 3.
To balance $\mathrm{N}, \mathrm{HNO}_3$ is multiplied by 10
$
4 \mathrm{Zn}+10 \mathrm{HNO} 3 \rightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{NH}_4 \mathrm{NO}_3+\mathrm{H}_3 \mathrm{O}
$
Step 4.
To balance oxygen, $\mathrm{H}_2 \mathrm{O}$ is multiplied by 3
$
4 \mathrm{Zn}+10 \mathrm{HNO}_3 \rightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{NH}_4 \mathrm{NO}_3+3 \mathrm{H}_3 \mathrm{O}
$
Question 10 .
Balance the following equation by ion-electron method In acidic medium.

Answer:
(i) $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_2 \mathrm{O}_4{ }^{2-}+\mathrm{SO}_2+\mathrm{I}^{-}$
Oxidation half reaction:

To balance, $\mathrm{SO}_2$ is added on RHS of the equation.
$
\mathrm{S}_2 \mathrm{O}_3{ }^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_2 \mathrm{O}_4{ }^{2-}+2 \mathrm{I}^{-}+\mathrm{SO}_2
$
To balance oxygen atom, $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ and $\mathrm{SO}_2$ is multiplied by 2 .
$
2 \mathrm{~S}_2 \mathrm{O}_3{ }^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_2 \mathrm{O}_4{ }^{2-}+2 \mathrm{I}-+2 \mathrm{SO}_2
$
(2) $\mathrm{Sb}^{3+}+\mathrm{MnO}_4 \rightarrow \mathrm{Sb}^{5+}+\mathrm{Mn}^{2+}$
Oxidation half reaction:
$
\mathrm{Sb}^{3+} \rightarrow \mathrm{Sb}^{5+}+2 \mathrm{e}^{-}
$
Reduction half reaction:

In equation (2), $\mathrm{H}_2 \mathrm{O}$ is added on L.HS to balance oxygen atom.
$
\mathrm{MnO}_2+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}
$
To balance Hydrogen atoms, $\mathrm{H}^{\prime}$ is added on RHS.
$
\mathrm{MnO}_4^{-}+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}
$
Equation (4) is multiplied by 2 and equation (I) is multiplied by 5 to equalise the electrons gained and electrons lost.

(3) $\mathrm{MnO}_4^{-}+\mathrm{I}^{-} \rightarrow \mathrm{MnO}_2+\mathrm{I}_2$
Oxidation half reaction:
$2 \mathrm{I}^{-}+2 \mathrm{e}^{-} \rightarrow \mathrm{I}^{-}$
Reduction half reaction:

To balance oxygen and hydrogen atoms, $\mathrm{H}^{+}$is added on RHS and $\mathrm{H}_2 \mathrm{O}$ is added on LHS.
$
2 \mathrm{MnO}_4+6 \mathrm{I}^{-}+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_2+2 \mathrm{I}_2+2 \mathrm{H}_2 \mathrm{O}
$
In acidic medium
(4) $\mathrm{MnO}_4^{-+} \mathrm{Fe}^{2+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}$
Oxidation half reaction:
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} \ldots \ldots .(1)$
Reduction half reaction:

To balance oxygen, $\mathrm{H}^{+}$is added on $\mathrm{RHS}$ and $\mathrm{H}_2 \mathrm{O}$ is added on LHS.
$
\mathrm{MnO}_4^{-+} 5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}
$
(5) $\mathrm{Cr}(\mathrm{OH})_4^{-}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{CrO}_4^{-}$
Oxidation half reaction:

Reduction half reaction:

To balance oxygen and hydrogen atoms, $\mathrm{OH}$ and $\mathrm{H}_2 \mathrm{O}$ are added.
$
2 \mathrm{Cr}(\mathrm{OH})_4+3 \mathrm{H}_2 \mathrm{O}_2+20 \mathrm{H}^{-} \rightarrow 2 \mathrm{CrO}_2^{-}+8 \mathrm{H}_2 \mathrm{O}
$
Question 11.
(a) Define equivalent mass of an oxidising agent.
(b) How would you calculate the equivalent mass of potassium permanganate?

Answer:
(a) The equivalent mass of an oxidizing agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation.
(b) Potassium permanganate is an oxidizing agent.

80 parts by mass of oxygen are given by $316 \mathrm{~g}$ of $\mathrm{KMnO}_4$.
$\therefore 8$ parts by mass of oxygen will be furnished by $\frac{316}{80} \times 8=31.6$
Equivalent mass of $\mathrm{KMnO}_4=31.6 \mathrm{~g} \mathrm{eq}^{-1}$.
Question 12.
(a) Define equivalent mass of an reducing agent.
(b) How would you determine the equivalent mass of Ferrous sulphate?
Answer:
(a) The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or one equivalent of any oxidising agent.
(b) Ferrous sulphate is a reducing agent.

16 parts by mass of oxygen oxidised 304 parts by mass of $\mathrm{FeSO}_4$.
$\therefore 8$ parts by mass of oxygen will oxidise $\frac{316}{80} \mathrm{x} 8$ parts by mass of ferrous sulphate $=152$.
The equivalent mass of ferrous sulphate (anhydrous) $=152$.
The equivalent mass of crystalline ferrous sulphate is $\left(\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}\right)=152+126=278$
The equivalent mass of crystalline ferrous sulphate $=278$.
Question 13.
A compound on analysis gave the following percentage composition: $\mathrm{C}=24.47 \%, \mathrm{H}=4.07 \%, \mathrm{Cl}=7$ $1.65 \%$. Find out its empirical formula.
Answer:

the empirical formula is $\mathrm{CH}_2 \mathrm{Cl}$.
Question 14.
A laboratory analysis of an organic compound gives the following mass percentage composition: $\mathrm{C}=60 \%$, $\mathrm{H}=4.48 \%$ and remaining oxygen.
Answer:

$\therefore$ the empirical formula is $\mathrm{C}_9 \mathrm{H}_8 \mathrm{O}_4$.
Question 15 .
An insecticide has the following percentage composition by mass: $47.5 \% \mathrm{C}, 2.54 \% \mathrm{H}$, and $50.0 \% \mathrm{Cl}$.
Determine its empirical formula and molecular formulae. Molar mass of the substance is $354.5 \mathrm{~g} \mathrm{~mol}^{-1}$
Answer:

The empirical formula is $\mathrm{C}_{14} \mathrm{H}_9 \mathrm{C}_{15}$.
Calculation of Molecular formula:
The empirical formula mass $\left(\mathrm{C}_{14} \mathrm{H}_9 \mathrm{C}_{15}\right)=(14 \mathrm{X} 12)+(9 \mathrm{X} 1)+(5 \mathrm{X} 35.5)$
Question 16.
An organic fruit smelling compound on analysis has the following composition by mass: $\mathrm{C}=54.54 \%, \mathrm{H}=$ $9.09 \%, O=36.36 \%$. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44 .
Answer:

$\therefore$ The empirical formula is $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$
Theempirical formula mass $\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)=(12 \times 2)+(1 \times 4)+(16 \times 1)=44$
Molecular mass $=2 \times$ Vapour density $=2 \times 44=88$
$\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{88}{44}=2$
Molecular formula $=(\text { Empirical formula })_{\mathrm{n}}$
$=\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)_2$
$\therefore$ Molecular formula $=\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2$
Question 17.
Calculate the percentage composition of the elements present in magnesium carbonate. How many Kg of
$\mathrm{CO}_2$ can be obtained from $100 \mathrm{Kg}$ of is $90 \%$ pure magnesium carbonate.
Molar mass of $\mathrm{MgCO}_3=84.32 \mathrm{~g} \mathrm{~mol}^{-1}$
Percentage of $\mathrm{Mg}=\frac{24}{84.32} \times 100=28.46 \%$
Percentage of $\mathrm{C}=\frac{12}{84.32} \times 100=14.23 \%$
Percentage of $\mathrm{O}_3=\frac{48}{84.32} \times 100=57.0 \%$

$84.32 \mathrm{~g}$ of $100 \%$ pure $\mathrm{MgCO}_3$ gives $44 \mathrm{~g}$ of $\mathrm{CO}_2$
$
\begin{aligned}
& \therefore 100 \times 10^3 \mathrm{~g} \text { of } 100 \% \text { pure } \mathrm{MgCO}_3 \text { gives }=\frac{44}{84.32} \times 100 \times 10^3 \\
& =52.182 \times 10^3 \mathrm{~g} \mathrm{CO}_2 \\
& 100 \% \text { pure } \mathrm{MgCO}_3 \text { gives } 52.182 \times 10^3 \mathrm{~g} \mathrm{CO}_2 \\
& \therefore 90 \% \text { pure } \mathrm{MgCO}_3 \text { will give } \frac{52.182 \times 10^3}{100} \times 90=46963.8 \mathrm{~g} \mathrm{CO}_2
\end{aligned}
$
Question 18.
Urea is prepared by the reaction between ammonia and carbon dioxide.

In one process, $637.2 \mathrm{~g}$ of $\mathrm{NH}_3$ are allowed to react with $1142 \mathrm{~g}$ of $\mathrm{CO}_2$.
(a) Which of the two reactants is the limiting reagent?
(b) Calculate the mass of $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}$ formed.
(c) how much of the excess reagent in grams is left at the end of the reaction?

Answer:

No. of moles of ammonia $=\frac{637.2}{17}=37.45$ mole
No. of moles of $\mathrm{CO}_2=\frac{1142}{44}=25.95$ moles
As per the balanced equation, one mole of $\mathrm{CO}_2$ requires 2 moles of ammonia.
$\therefore$ No. of moles of $\mathrm{NH}_3$ required to react with 25.95 moles of $\mathrm{CO}_2$ is $=\frac{2}{1} \times 25.95=51.90$ moles.
$\therefore 37.45$ moles of $\mathrm{NH}_3$ is not enough to completely react with $\mathrm{CO}_2$ (25.95 moles).
Hence, $\mathrm{NH}_3$ must be the limiting reagent, and $\mathrm{CO}_2$ is excess reagent.
(b) 2 moles of ammonia produce 1 mole of urea.
$\therefore$ Limiting reagent 37.45 moles of $\mathrm{NH}_3$ can produce $\frac{1}{2} \times 37.45$ moles of urea.
$=18.725$ moles of urea.
$\therefore$ The mass of 18.725 moles of urea $=$ No. of moles $\mathrm{x}$ Molar mass
$=18.725 \times 60$
$=1123.5 \mathrm{~g}$ of urea.
(c) 2 moles of ammonia requires 1 mole of $\mathrm{CO}_2$.
$\therefore$ Limiting reagent 37.45 moles of $\mathrm{NH}_3$ will require $\mathrm{x} 37.45$ moles of $\mathrm{CO}_2$.
$=18.725$ moles of $\mathrm{CO}_2$.
$\therefore$ No. of moles of the excess reagent $\left(\mathrm{CO}_2\right)$ left $=25.95-18.725=7.225$
The mass of the excess reagent $\left(\mathrm{CO}_2\right)$ left $=7.225 \times 44=317.9 \mathrm{~g} \mathrm{CO}_2$.

Question 19.
(a) Define oxidation number.
(b) What are the rules used to assign oxidation number?
Answer:
(a) Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely.
(b) Rules to assign oxidation number:
- Oxidation number of a substance in its elementary state is equal to zero $\left(\mathrm{H}_2, \mathrm{Br}_2, \mathrm{Na}\right)$
- Oxidation number of a mono-atomtic ion is equal to the charge on the Ton $\mathrm{Na}_4=+1$,
- Oxidation number of hydrogen in a compound is +1 (except hydrides).
- Oxidation number of hydrogen in metal hydrides is $-1\left(\mathrm{NaH}, \mathrm{CaH}_2\right)$.
- Oxidation number of oxygen in a compound is -2 (except $\mathrm{OF}_2$ and peroxides).
- Oxidation number of oxygen in peroxides is $-1(\mathrm{He}, \mathrm{Na} »)=,\mathrm{I}$.
- Oxidation number of oxygen in fluorinated compounds is either +1 or $+2 .\left(\mathrm{OF}_2=+2, \mathrm{O}_2 \mathrm{~F}_2+1\right)$.
- Fluorine has an oxidation number -1 in all its compounds.
- The sum of the oxidation number of all the atoms in neutral molecules is equal to Zero.
- For all ions, the sum of the oxidation number of all atoms is equal to the charge of the ion.
Question 20.
Balance the following equation by oxidation number method.
$
\mathrm{C}_6 \mathrm{H}_6+\mathrm{O}_2 \rightarrow \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$

(ii) Balance the changes in O.N. by multiplying the oxidant and reductant by suitable numbers $2 \mathrm{C}_6 \mathrm{H}_6+15 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
(iii) Balance the equation atomically (except $\mathrm{O}$ and $\mathrm{H}$ ).
$
2 \mathrm{C}_6 \mathrm{H}_6+15 \mathrm{O}_2 \rightarrow 12 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$
(iv) Balance $\mathrm{O}$ atoms by adding one $\mathrm{HzO}$ molecule to the RHS for making the number of molecules of $\mathrm{H}_2 \mathrm{O}$ to be 6 .
$
2 \mathrm{C}_6 \mathrm{H}_6+15 \mathrm{O}_2 \rightarrow 12 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}
$
Question 21.
Balance the following equation by oxidation number method.

(i) $\mathrm{KMnO}_4+\mathrm{HCl} \rightarrow \mathrm{KCl}+\mathrm{MnCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$

(ii) $2 \mathrm{KMnO}_4+1 \mathrm{OHCl} \rightarrow \mathrm{KCl}+\mathrm{MnCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$
(iii) Balance the equation atomically (except $\mathrm{O}$ and $\mathrm{H}$ ). $2 \mathrm{KMnO}_4+1 \mathrm{OHCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{MnCl}_2,+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$
(iv) Balance chlorine atoms by adding $\mathrm{HCl}$ and multiplying $\mathrm{Cl}_2$ by 5 . $2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{MnCl}_2+\mathrm{H}_2 \mathrm{O}+5 \mathrm{Cl}_2$
(v) To balance $\mathrm{O}$ and $\mathrm{H}, \mathrm{H}_2 \mathrm{O}$ is multiplied by 8 . $2 \mathrm{KMnO}_4+\mathrm{I} 6 \mathrm{HCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{MnCl}_4+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{Cl}_2$
Question 22 .
Balance the following equation by oxidation number method. $\mathrm{KMnO}_4+\mathrm{FeSO}_4+\mathrm{H} 2 \mathrm{SO}_4 \rightarrow \mathrm{K} 2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{Fe} 2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}$

Answer:

(ii) $2 \mathrm{KMnO}_4+10 \mathrm{FeSO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{Fe} 2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}$
(iii) Balance the equation atomically (except $\mathrm{O}$ and $\mathrm{H}$ ) and sulphate ions.
$
2 \mathrm{KMnO}_4+1 \mathrm{OFeSO}_4+8 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+5 \mathrm{Fe}_4\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}
$
(iv) Balance $\mathrm{O}$ atoms by multiplying $\mathrm{H}_2 \mathrm{O}$ by 8 .
$
2 \mathrm{KMnO}_4+1 \mathrm{OFeSO}_4+8 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+5 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+8 \mathrm{H}_2 \mathrm{O}
$
Question 23.
Balancing of molecular equation in alkaline medium.
$
\mathrm{MnO}_2+\mathrm{O}_2+\mathrm{KOH} \rightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{H}_2 \mathrm{O}
$

(ii) Balance the changes in $\mathrm{O}, \mathrm{N}$, by multiplying the oxidant and reductant by suitable numbers. $4 \mathrm{MnO}_2+2 \mathrm{O}_2+\mathrm{KOH} \rightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{H}_2 \mathrm{O}$
(iii) Balance the equation atomically (except $\mathrm{O}$ and $\mathrm{H}$ ).
$
4 \mathrm{MnO}_2+2 \mathrm{O}_2+\mathrm{KOH} \rightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{H}_2 \mathrm{O}
$
(iv) Balance oxygen and hydrogen atoms by multiplying $\mathrm{H} 20$ by 4 .
$
4 \mathrm{MnO}_2+2 \mathrm{O}_2+8 \mathrm{KOH} \rightarrow 4 \mathrm{~K}_2 \mathrm{MnO}_4+4 \mathrm{H}_2 \mathrm{O}
$
Question 24.
Explain the steps involved in ion-electron method for balancing redox reaction. Answer:
Ion electron method makes use of the Half reactions. Steps involved in this method are,
1. Write the equation in the net ionic form without attempting to balance it.
2. Write and locate the oxidation number of atoms undergoing oxidation and reduction from the knowledge of calculation of oxidation number.
3. Write two half reactions showing oxidation and reduction separately.
4. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
5. Add required number of $\mathrm{H}^{+}$ions to the side deficient in hydrogen atom if the reaction is in acidic medium
6. Add electrons to whichever side is necessary to make up the difference in oxidation number.
7. Add the two half reactions. The resulting equation is a net balanced equation.
8. For reactions in basic medium, add $\mathrm{H}_2 \mathrm{O}$ and hydrogen ion to balance $\mathrm{H}$ and $\mathrm{O}$.
9. Finally balance the equation by cancelling common species present on both sides of the equation.
Question 25 .
Write balanced equation for the oxidation of Ferrous ions to Ferric ions by permanganate ions in acid solution. The permanganate ion forms $\mathrm{Mn} 2+$ ions under these conditions.
Answer:
Net ionic reaction:
$\mathrm{MnO}_4^{-}+\mathrm{Fe}^{2+}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}$
Oxidation half reaction:
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}$
Reduction half reaction:
$
\mathrm{MnO}_4^{-}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}
$

To balance $\mathrm{O}$ and $\mathrm{H}, \mathrm{H}^{+}$and $\mathrm{H}_2 \mathrm{O}$ are added.
$
\mathrm{MnO}_4^{-+} 5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}
$
Question 26.
A flask A contains 0.5 mole of oxygen gas. Another flask B contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms.
Answer:
Flask A:
1 mole of oxygen gas $=6.023 \times 10^{23}$ molecules
$\therefore 0.5$ mole of oxygen gas $=6.023 \times 10^{23} \times 0.5$ molecules
The number of atoms in flask $\mathrm{A}=6.023 \times 10^{23} \times 0.5 \times 2=6.023 \times 10^{23}$ atoms.
Flask B:
1 mole of ozone gas $=6.023 \times 10^{23}$ molecules
0.4 mole of ozone gas $=6.023 \times 10^{23} \times 0.4$ molecules
The number of atoms in flask $B=6.023 \times 10^{23} \times 0.4 \times 3=7.227 \times 10^{23}$ atoms.
$\therefore$ Flask B contains a great number of oxygen atoms as compared to flask $A$.
Question 27.
(a) Formulate possible compounds of ' $\mathrm{Cl}$ ' in its oxidation state is:
$0,-1,+1,+3,+5,+7$
(b) $\mathrm{H}_2 \mathrm{O}_2$ act as an oxidising agent as well as reducing agent where as $\mathrm{O}_3$ act as only oxidizing agent.
Prove it.
(a) (1) $\mathrm{Cl}$ oxidation number $\mathrm{O}$ in $\mathrm{Cl}_2$.
(2) $\mathrm{Cl}$ oxidation number -1 in $\mathrm{HCl}$.
(3) $\mathrm{Cl}$ oxidation number +3 in $\mathrm{HClO}$. $_2$
(4) $\mathrm{Cl}$ oxidation number +5 in $\mathrm{KClO}_3$.
(5) $\mathrm{Cl}$ oxidation number +7 in $\mathrm{Cl}_2 \mathrm{O}_7$
(b) In $\mathrm{H}_2 \mathrm{O}_2$, oxidation number of oxygen is -1 and it can vary from $\mathrm{O}$ to -2 ( +2 is possible in $\mathrm{OF}_2$ ). The oxidation number can decrease or increase, because of this $\mathrm{H}_2 \mathrm{O}_2$ can act both oxidising and reducing agent. Ozone $\left(\mathrm{O}_3\right)$ only acts as oxidising agent since it decomposes to give nascent oxygen.
Question 28.
The $\mathrm{Mn}^{3+}$ ion is unstable in solution and undergoes disproportionation to give $\mathrm{Mn}^{2+}, \mathrm{MnO}_2$ and $\mathrm{H}$ ion. Write a balanced ionic equation for the reaction.
The skeletal equation is:
$
\mathrm{Mn}_{(\mathrm{aq})}^{3+} \rightarrow \mathrm{Mn}_{(\mathrm{aq})}^{2+}+\mathrm{MnO}_{2(\mathrm{aq})}+\mathrm{H}_{(\mathrm{aq})}^{+}
$
Oxidation half equation:

$
\mathrm{Mn}_{(\mathrm{aq})}^{+3} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}^{+4}
$
Balance O.N. by adding electrons,
$
\mathrm{Mn}^{3+}{ }_{(\mathrm{aq})} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{e}^{-}
$
Balance charge by adding $4 \mathrm{H}$ ions,
$
\mathrm{Mn}^{3+}{ }_{(\mathrm{aq})} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}+4 \mathrm{H}^{+}{ }_{(\mathrm{aq})}+\mathrm{e}^{-}
$
Balance $\mathrm{O}$ atoms by adding $2 \mathrm{H}_2 \mathrm{O}$
$
\mathrm{Mn}^{3+}{ }_{(\mathrm{aq})}+22 \mathrm{H}_2 \mathrm{O}_{\text {(1) }} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}+4 \mathrm{H}_{\text {(aq) }}^{+}+\mathrm{e}^{-}
$
Reduction half equation:
$
\mathrm{Mn}^{3+} \rightarrow \mathrm{Mn}^{+2+}
$
Balance $\mathrm{ON}$. by adding electrons:
$
\mathrm{Mn}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} \text { (aq) }
$
Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is
$
2 \mathrm{MH}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(1)} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+\mathrm{H}^{+}(\mathrm{aq})
$

Question 29.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for the reaction for this redox change taking place in water.
The skeletal equation is:

Reduction half equation:
$
\mathrm{Cl}_{2(\mathrm{aq})} \rightarrow \mathrm{Cl}_{\text {(aq) }}^{2-}
$
Balance $\mathrm{Cl}$ atoms,
$
\mathrm{Cl}_{2 \text { (aq) }} \rightarrow{ }^{-1} \mathrm{Cl}_{\text {(aq) }}^{-}
$
Balance O.N. by adding electrons
$
\mathrm{Cl}_{2(\mathrm{aq})}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}+2 \mathrm{e}^{-}
$
Oxidation half equation:

Balance $O, N$. by adding electrons
$
\mathrm{SO}_{2(\mathrm{aq})} \rightarrow \mathrm{SO}_4^{2-}(\mathrm{aq})+2 \mathrm{e}^{-}
$
Balance charge by adding $4 \mathrm{H}^{+}$ions:
$
\mathrm{SO}_{2(\mathrm{aq})} \rightarrow \mathrm{SO}_4^{2-}{ }_{(\mathrm{aq})}+4 \mathrm{H}_{(\mathrm{aq})}+2 \mathrm{e}^{-}
$
Balance $\mathrm{O}$ atoms by adding $2 \mathrm{H}_2 \mathrm{O}$
$
\mathrm{SO}_{2(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(1)} \rightarrow \mathrm{SO}_4^{2-}{ }_{(\mathrm{aq})}+4 \mathrm{H}_{\text {(aq) }}^{+}+2 \mathrm{e}^{-}
$
Adding equation (1) and (2), we have,
$
\mathrm{Cl}_{2(\mathrm{aq})}+\mathrm{SO}_{2(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(1)} \rightarrow \mathrm{Cl}_{2(\mathrm{aq})}+\mathrm{SO}_4^{2-} \text { (aq) }+4 \mathrm{H}^{+} \text {(aq) }
$
This represents the balanced redox reaction.

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Additional Questions - Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

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