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

Updated On May 15, 2024

By SaraNextGen

Basic Concepts of Chemistry and Chemical Calculations
Textual Evaluation Solved
I. Choose the Best Answer
Question 1.
$40 \mathrm{ml}$ of methane is completely burnt using $80 \mathrm{ml}$ of oxygen at room temperature. The volume of gas left after cooling to room temperature is
(a) $40 \mathrm{ml} \mathrm{CO} 2$ gas
(b) $40 \mathrm{ml} \mathrm{CO}_2$ gas and $80 \mathrm{ml} \mathrm{H}_2 \mathrm{O}$ gas
(c) $60 \mathrm{ml} \mathrm{CO}_2$ gas and $60 \mathrm{ml} \mathrm{H}_2 \mathrm{O}$ gas
(d) $120 \mathrm{ml} \mathrm{CO}_2$ gas
Answer:
(a) $40 \mathrm{ml} \mathrm{CO}$ gas
Solution:
$
\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(1)}
$

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct
Question 2.

An element $\mathrm{X}$ has the following isotopic composition ${ }^{200} \mathrm{X}=90 \%,{ }^{199} \mathrm{X}=8 \%$ and ${ }^{202} \mathrm{X}=2 \%$. The weighted average atomic mass of the element $\mathrm{X}$ is closest to ...........
(a) $201 \mathrm{u}$
(b) $202 \mathrm{u}$
(c) $199 \mathrm{u}$
(d) $200 \mathrm{u}$
Answer:
(d) $200 \mathrm{u}$
$
=\frac{(200 \times 90)+(199 \times 8)+(202 \times 2)}{100}=199.96=200 \mathrm{u}
$
Question 3.

Assertion:
Two mole of glucose contains $12.044 \times 10^{23}$ molecules of glucose.
Reason:
Total number of entities present in one mole of any substance is equal to $6.02 \times 10^{22}$
(a) both assertion and reason are true and the reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Correct reason:
Total number of entities present in one mole of any substance is equal to $6.022 \times 10^{23}$

Question 4.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) Both carbon and oxygen
(d) Neither carbon nor oxygen
Answer:
(b) Oxygen
Reaction 1:
$
2 \mathrm{C}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2
$
$2 \times 12$ g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon $=\frac{2 \times 12}{32} \times 8=6$
Reaction 2:
$
\mathrm{C}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2
$
$12 \mathrm{~g}$ carbon combines with $32 \mathrm{~g}$ of oxygen.
Hence, Equivalent mass of carbon $=\frac{12}{32} \times 8=3$
Question 5.
The equivalent mass of a trivalent metal element is $9 \mathrm{~g} \mathrm{eq}^{-1}$ the molar mass of its anhydrous oxide is
(a) $102 \mathrm{~g}$
(b) $27 \mathrm{~g}$
(c) $270 \mathrm{~g}$
(d) $78 \mathrm{~g}$
Answer:
(a) $102 \mathrm{~g}$
Let the trivalent metal be $\mathrm{M}^{3+}$
Equivalent mass $=$ mass of the metal $/$ valance factor
$9 \mathrm{~g} \mathrm{eq}^{-1}=$ mass of the metal $/ 3 \mathrm{eq}$
Mass of the metal $=27 \mathrm{~g}$
Oxide formed $\mathrm{M}_2 \mathrm{O}_3$
Mass of the oxide $=(2 \times 27)+(3 \times 16)=102 \mathrm{~g}$

Question 6.
The number of water molecules in a drop of water weighing $0.018 \mathrm{~g}$ is
(a) $6.022 \times 10^{26}$
(b) $6.022 \times 10^{23}$
(c) $6.022 \times 10^{20}$
(d) $99 \times 10^{22}$
Answer:
(c) $6.022 \times 10^{20}$
Weight of the water drop $=0.018 \mathrm{~g}$
No. of moles of water in the drop = Mass of water $/$ molar mass $=0.018 / 18=10^{-3} \mathrm{~mole}$
No of water molecules present in I mole of water $=6.022 \times 10^{23}$
"No. water molecules in one drop of water $(10 \mathrm{~mole})=6.022 \times 10^{23} \times 10^{-3}$ $=6.022 \times 10^{20}$
Question 7.
$1 \mathrm{~g}$ of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave $0.44 \mathrm{~g}$ of carbon dioxide gas. The percentage of impurity in the sample is
(a) $0 \%$
(b) $4.4 \%$
(c) $16 \%$
(d) $8.4 \%$
Answer:
(c) $16 \%$
$\mathrm{Mg} \mathrm{CO} 3 \rightarrow \mathrm{MgO}+\mathrm{CO}_2 \uparrow$
$\mathrm{Mg} \mathrm{CO}_3:(1 \times 24)+(1 \times 12)+(3 \times 16)=84 \mathrm{~g}$
$\mathrm{CO}_2:(1 \times 12)+(2 \times 16) 44 \mathrm{~g}$
$100 \%$ pure $84 \mathrm{~g} \mathrm{MgCO}_3$ on heating gives $44 \mathrm{~g} \mathrm{CO}_2$
Given that $\mathrm{I} g$ of $\mathrm{MgCO}_3$ on heating gives $0.44 \mathrm{~g} \mathrm{CO}_2$
Therefore, $84 \mathrm{~g} \mathrm{MgCO}_3$ sample on heating gives $36.96 \mathrm{~g} \mathrm{CO}_2=100 \%$
Percentage of purity of the sample $=\frac{100 \%}{44 \mathrm{gCO}_2} \times 36.96 \mathrm{~g} \mathrm{CO}_2=84 \%$
Percentage of impurity $=16 \%$
Question 8.
When $6.3 \mathrm{~g}$ of sodium bicarbonate is added to $30 \mathrm{~g}$ of acetic acid solution, the residual solution is found to weigh $33 \mathrm{~g}$. The number of moles of carbon dioxide released in the reaction is -
(a) 3
(b) 0.75
(c) 0.075
(a) 0.3
Answer:
(c) 0.075

The amount of $\mathrm{CO}_2$ released, $x=3.3 \mathrm{~g}$
No. of moles of $\mathrm{CO}_2$ released $=3.3 / 44=0.075 \mathrm{~mol}$
Question 9.
When 22.4 liters of $\mathrm{H}_2(\mathrm{~g})$ is mixed with 11.2 liters of $\mathrm{Cl}_2(\mathrm{~g})$, each at $273 \mathrm{~K}$ at $1 \mathrm{~atm}$ the moles of $\mathrm{HCl}(\mathrm{g})$, formed is equal to ..........
(a) 2 moles of $\mathrm{HCl}(\mathrm{g})$
(b) 0.5 moles of $\mathrm{HCl}(\mathrm{g})$
(c) 1.5 moles of $\mathrm{HCl}(\mathrm{g})$
(d) 1 moles of $\mathrm{HCl}(\mathrm{g})$
Answer:
(d) 1 moles of $\mathrm{HCl}$ (g)
Solution:
$
\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})
$

Question 10.
Hot concentrated sulfuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior?
(a) $\mathrm{Cu}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{C}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 4 \mathrm{CO}_2+2 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+2 \mathrm{HCl}$
(d) none of the above
Answer:
(c) $\mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+2 \mathrm{HCl}$

Question 11.
Choose the disproportional reaction among the following redox reactions.
(a) $3 \mathrm{Mg}$ (s) $+\mathrm{N}_2$ (g) $\rightarrow \mathrm{Mg}_2 \mathrm{~N}_2$ (s)
(b) $\mathrm{P}_4$ (s) $+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3$ (g) $+3 \mathrm{NaH}_2 \mathrm{PO}_2$ (aq)
(c) $\mathrm{Cl}_2$ (g) $+2 \mathrm{Kl}(\mathrm{aq}) \rightarrow 2 \mathrm{KCl}$ (aq) $+\mathrm{I}_2$
(d) $\mathrm{Cr}_2 \mathrm{O}_3$ (s) $+2 \mathrm{Al}$ (s) $\rightarrow \mathrm{A}_2 \mathrm{O}_3$ (s) $+2 \mathrm{Cr}$ (s)
Answer:
(b) $\mathrm{P}_4$ (s) $+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3$ (g) $+3 \mathrm{NaH}_2 \mathrm{PO}_2$ (aq)
$
\mathrm{P}_4(\mathrm{~s})+3 \mathrm{NaOH}+3 \mathrm{H}_2 \rightarrow \mathrm{PH}_3(\mathrm{~g})+3 \mathrm{NaH}_2 \mathrm{PO}_2(\mathrm{aq})
$
Question 12.

The equivalent mass of potassium permanganate in alkaline medium is
$
\mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-}
$
(a) 31.6
(b) 52.7
(c) 79
(d) None of these
Answer:
(b) 52.7
The reduction reaction of the oxidizing agent $\left(\mathrm{MnO}_4\right)$ involves gain of 3 electrons.
Hence the equivalent mass $=\left(\right.$ Molar mass of $\left.\mathrm{KMnO}_4\right) / 3=158.1 / 3=52.7$

Question 13 .
Which one of the following represents $180 \mathrm{~g}$ of water?
(a) 5 Moles of water
(b) 90 moles of water
(c) $\frac{6.022 \times 10^{23}}{180}$ Molecules of water
(d) $6.022 \times 10^{24}$ Molecules of water
Answer:
(d) $6.022 \times 10^{24}$ Molecules of water
No. of moles of water present in $180 \mathrm{~g}$
$=$ Mass of water / Molar mass of water
$=180 \mathrm{~g} / 18 \mathrm{~g} \mathrm{~mol}^{-1}=10$ moles
One mole of water contains
$=6.022 \times 10^{23}$ water molecules
10 mole of water contains $=6.022 \times 10^{23} \times 10$
$=6.022 \times 10^{24}$ water molecules
Question 14.
$7.5 \mathrm{~g}$ of a gas occupies a volume of 5.6 liters at $0^{\circ} \mathrm{C}$ and 1 atm pressure. The gas is
(a) $\mathrm{NO}$
(b) $\mathrm{N}_2 \mathrm{O}$
(c) $\mathrm{CO}$
(d) $\mathrm{CO}_2$
Answer:
(a) $\mathrm{NO}$
$7.5 \mathrm{~g}$ of gas occupies a volume of 5.6 liters at $273 \mathrm{~K}$ and 1 atm pressure Therefore, the mass of gas that occupies a volume of 22.4 liters -
$\frac{7.5 g}{5.6 L} \times 22.4 \mathrm{~L}=30 \mathrm{~g}$
Molar mass of NO $(14+16)=30 \mathrm{~g}$
Question 15.
Total number of electrons present in $1.7 \mathrm{~g}$ of ammonia is
(a) $6.022 \times 10^{23}$
(b) $\frac{6.022 \times 10^{22}}{1.7}$
(c) $\frac{6.022 \times 10^{24}}{1.7}$
(d) $\frac{6.022 \times 10^{23}}{1.7}$
Answer:
(a) $6.022 \times 10^{23}$

No. of electrons present in one ammonia $\left(\mathrm{NH}_3\right)$ molecule $(7+3)=10$
No. of moles of ammonia $=\frac{\text { Mass }}{\text { Molarmass }}$ $=\frac{1.7 \mathrm{~g}}{17 \mathrm{gmol}^{-1}}=0.1 \mathrm{~mol}$
No. of molecules present in 0ne ammonia $=0.1 \times 6.022 \times 10^{23}=6 . \mathrm{O} 22 \times 10^{22}$
No. of electrons present in $0.1 \mathrm{~mol}$ of ammonia $10 \times 6.022 \times 10^{22}=6.022 \times 10^{23}$
Question 16.
The correct increasing order of the oxidation state of sulphur in the anions $\mathrm{SO}_4{ }^{2-}, \mathrm{SO}_3{ }^{2-}, \mathrm{S}_2 \mathrm{O}_4{ }^{2-}, \mathrm{S}_2 \mathrm{O}_6{ }^{2-}$ is
(a) $\mathrm{SO}_3{ }^{2-}<\mathrm{SO}_3{ }^{2-}<\mathrm{S}_2 \mathrm{O}_4{ }^{2-}<\mathrm{S}_2 \mathrm{O}_6{ }^{2-}$
(c) $\mathrm{S}_2 \mathrm{O}_4{ }^{2-}<\mathrm{SO}_3{ }^{2-}<\mathrm{S}_2 \mathrm{O}_6{ }^{2-}<\mathrm{SO}_4{ }^{2-}$
(d) $\mathrm{S}_2 \mathrm{O}_6^{2-}<\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_4{ }^{2-}<\mathrm{SO}_3^{2-}$
Answer:
(c) $\mathrm{S}_2 \mathrm{O}_4{ }^{2-}<\mathrm{SO}_3{ }^{2-}<\mathrm{S}_2 \mathrm{O}_6{ }^{2-}<\mathrm{SO}_4{ }^{2-}$

Question 17.

The equivalent mass of ferrous oxalate is
(a) $\frac{\text { molar mass of ferrous oxalate }}{1}$
(b) $\frac{\text { molar mass of ferrous oxalate }}{2}$
(c) $\frac{\text { molar mass of ferrous oxalate }}{3}$
(d) none of these
Answer:
(c) $\frac{\text { molar mass of ferrous oxalate }}{3}$

Question 18.
If Avogadro number were changed from $6.022 \times 10^{23}$ to $6.022 \times 10^{20}$, this would change
(a) the ratio of chemical species to each other in a balanced equation
(b) the ratio of elements to each other in a compound
(c) the definition of mass in units of grams
(d) the mass of one mole of carbon

Answer:
(d) the mass of one mole of carbon
Question 19.
Two 22.4 liter containers $\mathrm{A}$ and $\mathrm{B}$ contains $8 \mathrm{~g}$ of $\mathrm{O}_2$ and $8 \mathrm{~g}$ of $\mathrm{SO}_2$ respectively, at $273 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure, then
(a) number of molecules in $\mathrm{A}$ and $\mathrm{B}$ are same
(b) number of molecules in $\mathrm{B}$ is more than that in $\mathrm{A}$
(c) the ratio between the number of molecules in $\mathrm{A}$ to the number of molecules in $\mathrm{B}$ is $2: 1$
(d) number of molecules in $\mathrm{B}$ is three times greater than the number of molecules in $\mathrm{A}$
Answer:
(c) The ratio between the number of molecules in $\mathrm{A}$ to number of molecules in $\mathrm{B}$ is $2: 1$
Question 20

. $1.865 \%$ potassium chloride solution?
(a) $3.59 \mathrm{~g}$
(b) $7 \mathrm{~g}$
(c) $14 \mathrm{~g}$
(d) $28 \mathrm{~g}$
Answer:
(a) $3.59 \mathrm{~g}$
$\mathrm{AgNO}_3+\mathrm{KCl} \rightarrow \mathrm{KNO}_3+\mathrm{AgCl}$
Solution:
$50 \mathrm{~mL}$ of $8.5 \%$ solution contains $4.25 \mathrm{~g}$ of $\mathrm{AgNO}_3$
No. of moles of $\mathrm{AgNO}_3$ present in $50 \mathrm{~mL}$ of $8.5 \% \mathrm{AgNO}_3$ solution
$=$ Mass $/$ Molar mass $=4.25 / 170=0.025$ moles
Similarly, No of moles of $\mathrm{KCl}$ present in loo $\mathrm{mL}$ of $1.865 \% \mathrm{KCl}$ solution $=1.865 / 74.5=0.025$ moles
So total amount of $\mathrm{AgCl}$ formed is 0.025 moles (based on the stoichiometry calculator)
Amount of $\mathrm{AgCl}$ present in 0.025 moles of $\mathrm{AgCl}$
$=$ no. of moles $\times$ molar mass
$=0.025 \times 143.5=3.59 \mathrm{~g}$

Question 21.
The mass of a gas that occupies a volume of $612.5 \mathrm{ml}$ at room temperature and pressure $\left(25^{\circ} \mathrm{C}\right.$ and $1 \mathrm{~atm}$ pressure) is $1.1 \mathrm{~g}$. The molar mass of the gas is
(a) $66.25 \mathrm{~g} \mathrm{~mol}^{-1}$
(b) $44 \mathrm{~g} \mathrm{~mol}^{-1}$
(c) $24.5 \mathrm{~g} \mathrm{~mol}^{-1}$
(d) $662.5 \mathrm{~g} \mathrm{~mol}^{-1}$
Answer:
(b) $44 \mathrm{~g} \mathrm{~mol}^{-1}$
Solution:
No. of moles of a gas that occupies a volume of $612.5 \mathrm{ml}$ at room temperature and pressure $\left(25^{\circ} \mathrm{C}\right.$ and $1 \mathrm{~atm}$ pressure)
$
=612.5 \times 10^{-3} \mathrm{~L} / 24.5 \mathrm{~L} \mathrm{~mol}^{-1}
$
$=0.025$ moles
We know that,
Molar mass = Mass $/$ no of moles
$=1.1 \mathrm{~g} / 0.025 \mathrm{~mol}=44 \mathrm{~g} \mathrm{~mol}^{-1}$
Question 22.
Which of the following contain same number of carbon atoms as in $6 \mathrm{~g}$ of carbon -12 ?
(a) 7.5 g ethane
(b) 8 g methane
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
No. of moles of carbon present in $6 \mathrm{~g}$ of C $-12=$ Mass / Molar mass
$=6 / 12=0.5$ moles $=0.5 \times 6.022 \times 10^{23}$ carbon atoms. .
No. of moles in $8 \mathrm{~g}$ of methane $=8116=0.5$ moles
$=0.5 \times 6.022 \times 10^{23}$ carbon atoms.
No. of moles in $7.5 \mathrm{~g}$ of ethane $=7.5 / 16=0.25$ moles
$=2 \times 0.25 \times 6.022 \times 10^{23}$ carbon atoms.

Question 23.
Which of the following compound(s) has/have percentage of carbon same as that in ethylene $\left(\mathrm{C}_2 \mathrm{H}_4\right)$ ?
(a) propene
(b) ethyne
(c) benzene
(d) ethane
Answer:
(a) propene
Solution:
Molar mass of carbon
Percentage of carbon in ethylene $\left(\mathrm{C}_2 \mathrm{H}_6\right)=\frac{\text { Molar mass of carbon }}{\text { Molar mass }} \times 100$ $=\frac{24}{28} \times 100=85.71 \%$
Percentage of carbon in propene $\left(\mathrm{C}_3 \mathrm{H}_6\right)=\frac{24}{28} \times 100=85.71 \%$
Question 24 .
Which of the following is/are true with respect to carbon - 12 ?
(a) relative atomic mass is $12 \mathrm{u}$
(b) oxidation number of carbon is +4 in all its compounds.
(c) I mole of carbon -12 contain $6.022 \times 10^{22}$ carbon atoms.
(d) all of these
Answer:
(a) relative atomic mass is $12 \mathrm{u}$
Question 25 .
Which one of the following is used as a standard for atomic mass?
(a) ${ }_6 \mathrm{C}^{12}$
(b) ${ }_7 \mathrm{C}^{12}$
(c) ${ }_6 \mathrm{C}^{13}$
(d) ${ }_6 \mathrm{C}^{14}$

Answer:
(a) ${ }_6 \mathrm{C}^{12}$
II. Write brief answer to the following questions
Question 26.
Define relative atomic mass.
On the basis of carbon, the relative atomic mass of element is defined as the ratio of mass of one atom of the element to the mass of $1 / 12$ th mass of one atom of Carbon -12 .
$
\text { Relative atomic mass }=\frac{\text { Mass of one atom of the element }}{\text { Mass of } 1 / 12^{\text {th }} \text { mass of one atom of Carbon- } 12}
$
Question 27.
What do you understand by the term mole?
Answer:
The mole is defined as the amount of a substance which contains $6.023 \times 10^{23}$ particles such as atoms, molecules or ions. It is represented by the symbol
Question 28.
Define equivalent mass.
Answer:
The equivalent mass of an element is the number of parts of the mass of an element which combines with or displaces 1.008 parts of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine.
Question 29.
What do you understand by the term oxidation number?
Answer:
Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely. The oxidation numbers reflect the number of electrons "transferred".
Question 30 .
Distinguish between oxidation and reduction.
Answer:
Oxidation:
According to the classical concept, oxidation is a process of addition of oxygen or removal of hydrogen.
Removal of hydrogen
$
2 \mathrm{H}_2 \mathrm{~S}+\mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}+2 \mathrm{~S}
$
Addition of oxygen
$
\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2
$
According to the electronic concept, loss of electrons is called oxidation reaction.
$
\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}
$
During oxidation, oxidation number increases.
Dining oxidation, reducing agent gets oxidised.

Reduction:
Reduction is a process of removal of oxygen or addition of hydrogen.
Addition of hydrogen
$
\mathrm{Ca}+\mathrm{H}_2 \rightarrow \mathrm{CaH}_2
$
Removal of oxygen
$
\mathrm{ZnO}+\mathrm{C} \rightarrow \mathrm{Zn}+\mathrm{CO}
$
According to the electronic concept, gain of electrons is called reduction reaction.
$
\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}
$
During reduction, oxidation number decreases.
During reduction, oxidising agent gets reduced.
Question 31.
Calculate the molar mass of the following compounds.
1. urea $\left[\mathrm{CO}\left(\mathrm{NH}_2\right)_2\right]$
2. acetone $\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]$
3. boric acid $\left[\mathrm{H}_3 \mathrm{BO}_3\right]$
4. sulphuric acid $\left[\mathrm{H}_2 \mathrm{SO}_4\right]$
Answer:
1. urea $\left[\mathrm{CO}\left(\mathrm{NH}_2\right)_2\right]$
Atomic mass of $\mathrm{C}=12$
Atomic mass of $\mathrm{O}=16$
Atomic mass of $2(\mathrm{~N})=28$
Atomic mass of $4(\mathrm{H})=4$
$\therefore$ Molar mass of Urea $=60$
2. Acetone $\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]$
Atomic mass of $3(C)=36$
Atomic mass of $1(0)=16$
Atomic mass of $6(\mathrm{H})=6$
$\therefore$ Molar mass of Acetone $=58$

3. Boric acid $\left[\mathrm{H}_3 \mathrm{BO}_3\right]$
Atomic mass of $\mathrm{B}=10$
Atomic mass of $3(\mathrm{H})=3$
Atomic mass of $3(O)=48$
$\therefore$ Molar mass of Boric acid $=61$
4. Sulphuric acid ${ }_2\left[\mathrm{H}_2 \mathrm{SO}_4\right]$
Atomic mass of $2(\mathrm{H})=2$
Atomic mass of $1(\mathrm{~S})=32$
Atomic mass of $4(\mathrm{O})=64$
$\therefore$ Molar mass of Sulphuric acid $=98$
Question 32 .
The density of carbon dioxide is equal to $1.977 \mathrm{~kg} \mathrm{~m}^{-3}$ at $273 \mathrm{~K}$ and 1 atm pressure. Calculate the molar mass of $\mathrm{CO}_2$
Answer:
Molecular mass $=$ Density $\times$ Molar volume
Molar volume of $\mathrm{CO}_2=2.24 \times 10^{-2} \mathrm{~m}^3$
Density of $\mathrm{CO}_2=1.977 \mathrm{~kg} \mathrm{~m}^{-3}$
Molecular mass of $\mathrm{CO}_2=1.977 \times 10^3 \mathrm{gm}^{-3} \times 2.24 \times 10^{-2} \mathrm{~m}^3$ $=1.977 \times 10^{-1} \times 2.24=44 \mathrm{~g}$
Question 33.
Which contains the greatest number of moles of oxygen atoms?
1. $1 \mathrm{~mol}$ of ethanol
2. $1 \mathrm{~mol}$ of formic acid
3. $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$
Answer:
1. $1 \mathrm{~mol}$ of ethanol $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ (ethanol) - Molar mass $=24+6+16=46$
$46 \mathrm{~g}$ of ethanol contains $1 \times 6.023 \times 10^{23}$ number of oxygen atoms.

2. $1 \mathrm{~mol}$ of formic acid.
$\mathrm{HCOOH}$ (formic acid) - Molar mass $=2+12+32=46$
$46 \mathrm{~g}$ of $\mathrm{HCOOH}$ contains $2 \times 6.023 \times 10^{23}$ number of oxygen atoms.
3. $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$
$\mathrm{H}_2 \mathrm{O}$ (water) - Molar mass $=2+16=18$
$18 \mathrm{~g}$ of water contains $1 \times 6.023 \times 10^{23}$ number of oxygen atoms.
$\therefore 1$ mole of formic acid contains the greatest number of oxygen atoms.
Question 34.
Calculate the average atomic mass of naturally occurring magnesium using the following data

Answer:
Isotopes of $\mathrm{Mg}$.
Atomic mass $=\mathrm{Mg}^{24}=23.99 \times \frac{783.99}{100}=18.95$
Atomic mass $=\mathrm{Mg}^{26}=24.99 \times \frac{10}{100}=2.499$
Atomic mass $=\mathrm{Mg}^{25}=25.98 \times \frac{11.01}{100}=2.860$
Average Atomic mass $=24.309$
Average atomic mass of $\mathrm{Mg}=24.309$
Question 35.
In a reaction $x+y+z_2 \rightarrow \mathrm{xyz}_2$, identify the limiting reagent if any, in the following reaction mixtures.
(a) 200 atoms of $x+200$ atoms of $y+50$ molecules of $z_2$
(b) $1 \mathrm{~mol}$ of $\mathrm{x}+1 \mathrm{~mol}$ of $\mathrm{y}+3 \mathrm{~mol}$ of $\mathrm{z}_2$
(c) 50 atoms of $x+25$ atoms of $y+50$ molecules of $z_2$
(d) $2.5 \mathrm{~mol}$ of $\mathrm{x}+5 \mathrm{~mol}$ of $\mathrm{y}+5 \mathrm{~mol}$ of $\mathrm{z}_2$
Answer:
$
\mathrm{x}+\mathrm{y}+\mathrm{z}_2
$
(a) 200 atoms of $x+200$ atoms of $y+50$ molecules of $z_2$ According to the reaction, 1 atom of $x$ reacts with one atom of $y$ and one molecule of $z$ to give product. In the case (a) 200 atoms of $x, 200$ atoms of $y$ react with 50 molecules of $z_2$ (4 part) i.e. 50 molecules of $z_2$ react with 50 atoms of $x$ and 50 atoms of $y$. Hence $\mathrm{z}$ is the limiting reagent.
(b) $1 \mathrm{~mol}$ of $\mathrm{x}+1 \mathrm{~mol}$ of $\mathrm{y}+3 \mathrm{~mol}$ of $\mathrm{z}_2$
According to the equation 1 mole of $z_2$ only react with one mole of $\mathrm{x}$ and one mole of $\mathrm{y}$. If 3 moles of $z_2$ are there, $z$ is limiting reagent.
(c) 50 atoms of $x+25$ atoms of $y+50$ molecules of $z_2$
25 atoms of $\mathrm{y}$ react with 25 atoms of $\mathrm{x}$ and 25 molecules of $\mathrm{z}_2$. So $\mathrm{y}$ is the limiting reagent.
(d) $2.5 \mathrm{~mol}$ of $\mathrm{x}+5 \mathrm{~mol}$ of $\mathrm{y}+5 \mathrm{~mol}$ of $\mathrm{z}_2$
$2.5 \mathrm{~mol}$ of $\mathrm{x}$ react with 2.5 mole of $\mathrm{y}$ and 2.5 mole of $\mathrm{z}_2$. So $\mathrm{x}$ is the limiting reagent.

Question 36.
Mass of one atom of an element is $6.645 \times 10^{-23} \mathrm{~g}$. How many moles of element are there in $0.320 \mathrm{~kg}$ ?
Answer:
Mass of one atom of an element $=6.645 \times 10^{-23} \mathrm{~g}=$ Atomic mass.
Mass of given element $=0.320 \mathrm{~kg}$
Number of moles $=$
Atomic mass
$
\begin{aligned}
& \text { Number of moles }=\frac{\text { Mass }}{\text { Atomic mass }} \\
&=\frac{0.320 \times 1000 \mathrm{~g}}{6.645 \times 10^{-23} \mathrm{~g}} \\
&=\frac{320 \times 10^{23}}{6.645} \\
&=48.156 \times 10^{23} \\
&=4.8156 \times 10^{24} \mathrm{moles} . \\
&=48.156 \times 10^{-23} \\
&=4.8156 \times 10^{-24} \text { moles. }
\end{aligned}
$
Question 37.
What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
Answer:
Molecular mass:
- Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.
- It can be calculated by adding the relative atomic masses of its constituent atoms.
- For carbon monoxide (CO) Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 $+16=28 \mathrm{u}$.

Molar mass:
- It is defined as the mass of one mole of a substance.
- The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in $\mathrm{g} \mathrm{mol}^{-1}$.
- For carbon monoxide (CO) $12+16=28 \mathrm{~g} \mathrm{~mol}^{-1}$ Both molecular mass and molar mass are numerically same but the units are different.
Question 38 .
What is the empirical formula of the following?
1. Fructose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ found in honey
2. Caffeine $\left(\mathrm{C}_8 \mathrm{H}_{10} \mathrm{~N}_4 \mathrm{O}_2\right)$ a substance found in tea and coffee.
Answer:
1. Fructose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$
Empirical formula is the simplest formula. So it is divided by 6 and so empirical formula is $\mathrm{CH}_2 \mathrm{O}$.
2. Caffeine $\left(\mathrm{C}_8 \mathrm{H}_{10} \mathrm{~N}_4 \mathrm{O}_2\right)$
Simplified formula $=\frac{\text { molecular formula }}{2}$
Empirical formula $=\mathrm{C}_4 \mathrm{H}_5 \mathrm{~N}_2 \mathrm{O}$.
Question 39.
The reaction between aluminium and ferric oxide can generate temperatures up to $3273 \mathrm{~K}$ and is used in welding metals. (Atomic mass of $\mathrm{AC}=21 \mathrm{u}$ Atomic mass of $0=16 \mathrm{u}$ ) $2 \mathrm{Al}+\mathrm{Fe}_2 \mathrm{O}_2 \rightarrow \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Fe}$; If, in this process, $324 \mathrm{~g}$ of aluminium is allowed to react with $1.12 \mathrm{~kg}$ of ferric oxide.
1. Calculate the mass of $\mathrm{Al}_2 \mathrm{O}_3$ formed.
2. How much of the excess reagent is left at the end of the reaction?
Answer:
(i) $2 \mathrm{Al}+\mathrm{Fe}_2 \mathrm{O}_3 \rightarrow \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Fe}$
$54 \mathrm{~g} \quad 160 \mathrm{~g} \quad 102 \mathrm{~g} \quad 112 \mathrm{~g}$
1. As per balanced equation $54 \mathrm{~g} \mathrm{~A} 1$ is required for $112 \mathrm{~g}$ of iron and $102 \mathrm{~g}$ of $\mathrm{Al}_2 \mathrm{O}_3$.
$54 \mathrm{~g}$ of $\mathrm{Al}$ gives $102 \mathrm{~g}$ of $\mathrm{Al}_2 \mathrm{O}_3$.
$\therefore 324 \mathrm{~g}$ of $\mathrm{Al}$ will give $\frac{102}{54} \times 324=612 \mathrm{~g}$ of $\mathrm{Al}_2 \mathrm{O}_3$.
2. $54 \mathrm{~g}$ of $\mathrm{Al}$ requires $160 \mathrm{~g}$ of $\mathrm{Fe}_2 \mathrm{O}_3$ for welding reaction.
$\therefore 324 \mathrm{~g}$ of Al will require $\frac{160}{54} \times 324=960 \mathrm{~g}$ of $\mathrm{Fe}_2 \mathrm{O}_3$.
$\therefore$ Excess $\mathrm{Fe}_2 \mathrm{O}_3-\mathrm{Un}$ reacted $\mathrm{Fe}_2 \mathrm{O}_3=1120-960=160 \mathrm{~g}$
$160 \mathrm{~g}$ of excess reagent is left at the end of the reaction.
Question 40 .
How many moles of ethane is required to produce $44 \mathrm{~g}$ of $\mathrm{CO}_2(\mathrm{~g})$ after combustion.

Answer:
$
\mathrm{C}_2 \mathrm{H}_6+3 \frac{1}{2} \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}
$
Ethane Carbon dioxide
1 mole of ethane $\stackrel{\text { Combustion }}{\longrightarrow} 2$ moles of $\mathrm{CO}_2$
$\therefore 44 \mathrm{~g}$ of $\mathrm{CO}_2=\mathrm{I}$ mole of $\mathrm{CO}_2$
2 moles of $\mathrm{CO}_2$ is produced by 1 mole of ethane.
$\therefore 1$ mole of $\mathrm{CO}_2$ will be produced by $=$ ?
$\therefore$ To produce 1 mole of $\mathrm{CO}_2$, the required mole of ethane is $=\frac{1}{2} \times 1=0.5$ mole of ethane.
Question 41.
Hydrogen peroxide is an oxidizing agent. It dioxides ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:
$\mathrm{H}_2 \mathrm{O}_2$-Oxidizing agent
$
\mathrm{Fe}^{2+}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O} \text { (Acetic medium) }
$
Ferrous ion is oxidized by $\mathrm{H}_2 \mathrm{O}_2$ to Ferric ion.
The balanced equation is $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} \mathrm{x} 2$
$
\begin{aligned}
& \mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \\
& 2 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \\
& \mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \\
& \overline{2 \mathrm{Fe}^{2+}}+\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+} \rightarrow \overline{2 \mathrm{Fe}^{3+}+2 \mathrm{H}_2 \mathrm{O}} \\
&
\end{aligned}
$
Question 42.
Calculate the empirical and molecular formula of a compound containing $76.6 \%$ carbon, $6.38 \%$ hydrogen and rest oxygen its vapour density is 47.
Answer:

Empirical formula $=\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}$
Va-pour density 47
$\therefore$ Molecular mass $=2 \mathrm{x}$ vapor density $=2 \times 47=94$
Molecular formula Empirical formula $\times \mathrm{n}$
Molecular mass $\mathrm{x} n$
$\mathrm{n}=\frac{\text { Molecularmass }}{\text { Empirical formulamass }}=\frac{94}{94}=1$
$\therefore$ Molecular formula $=\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}$
Question 43.
$\mathrm{A}$ Compound on analysis gave $\mathrm{Na}=14.31 \% \mathrm{~S}=9.97 \% \mathrm{H}=6.22 \%$ and $\mathrm{O}=69.5 \%$ calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
Answer:

All $\mathrm{H}$ combines with 10 oxygen atoms to form as $10 \mathrm{H}_2 \mathrm{O}$.
So the empirical formula is $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}$
Empirical formula mass $=(23 \times 2)+(32 \times 1)+(16 \times 4)+(10 \times 18)$
$=46+32+64+180=322$
$\mathrm{n}=\frac{\text { Molecularmass }}{\text { Empirical formulamass }}=\frac{322}{322}=1$
Molecular formula $=\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}$
Question 44.
Balance the following equations by oxidation number method
1. $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{I}_2+\mathrm{H}_2 \mathrm{O}$
2. $\mathrm{KMnO}_4+\mathrm{Na}_2 \mathrm{SO}_3 \rightarrow \mathrm{MnO}_2+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{KOH}$
3. $\mathrm{Cu}+\mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
4. $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
Answer:
1. $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_2+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{I}_2+\mathrm{H}_2 \mathrm{O}$
Step - 1

Step -2 $\mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7+6 \mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+3 \mathrm{I}_2+\mathrm{H}_2 \mathrm{O}$
Step -3
To balance other atoms
$
\begin{aligned}
& \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+6 \mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 4 \mathrm{~K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+3 \mathrm{I}_2+\mathrm{H}_2 \mathrm{O} \\
& \text { Step }-4 \\
& \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7+6 \mathrm{KI}+7 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 4 \mathrm{~K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O} \\
& \text { 2. } \mathrm{KMnO}_4+\mathrm{Na}_2 \mathrm{SO}_3 \rightarrow \mathrm{MnO}_2+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{KOH} \text { (Alkaline medium) } \\
& \text { Step }-1
\end{aligned}
$

Step -2
$
2 \mathrm{KMnO}_4+3 \mathrm{Na}_2 \mathrm{SO}_3 \rightarrow 2 \mathrm{MnO}_2+3 \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{KOH}
$
Step -3
balancing potassium, $\mathrm{KOH}$ is multiplied by 2
$
2 \mathrm{KMnO}_4+3 \mathrm{Na}_2 \mathrm{SO}_3 \rightarrow 2 \mathrm{MnO}_2+3 \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{KOH}
$
Step -4
To balance $\mathrm{H}$ atom, $\mathrm{H}_2 0$ is added on reactant side.
$
2 \mathrm{KMnO}_4+3 \mathrm{Na}_2 \mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MnO}_2+3 \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{KOH}
$
3. $\mathrm{Cu}+\mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
Step -1

Step -2 $\mathrm{Cu}+\mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
Step -3
To balance Nitrogen, $2 \mathrm{HNO}_3$ is multiplied by 2 and $\mathrm{NO}_2$ is multiplied by 2
$\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$
Step 4.
To balance oxygen, $\mathrm{H}_2 \mathrm{O}$ is multiplied by 2
$\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}$
4. $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
Step -1

Step -2
$5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+10 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
Step -3
To balance $\mathrm{K}, \mathrm{KMnO}_4$ and $\mathrm{MnSO}_4$ are multiplied by 2
$
5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$
Step -4
To balance $\mathrm{O}$ and $\mathrm{H}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{SO}_4$ are multiplied by 3 and 6 .
$
5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}
$
Question 45.
Balance the following equations by ion electron method.
1. $\mathrm{KMnO}_4+\mathrm{SnCl}_2+\mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{SnCl}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{KCl}$
2. $\mathrm{C}_2 \mathrm{O}_4{ }^{2-}+\mathrm{Cr}_2 \mathrm{O}_{72-} \rightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_2$ (in acid medium)
3. $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+\mathrm{NaI}$ (in acid medium)
4. $\mathrm{Zn}+\mathrm{NO}_3^{-} \rightarrow \mathrm{Zn}^{2+}+\mathrm{NO}$
Answer:
1. $\mathrm{KMnO}_4+\mathrm{SnCl}_2+\mathrm{HCl} \rightarrow \mathrm{MnCl}_3+\mathrm{SnCl}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{KCl}$
Oxidation half reaction: (loss of electrons)

Reduction half reaction: (gain of electrons)

Add $\mathrm{H}_2 \mathrm{O}$ to balance oxygen atoms.

Add $\mathrm{HCl}$ to balance hydrogen atoms
$
\mathrm{KMnO}_4+5 \mathrm{e}_{-}+8 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+4 \mathrm{H}_2 \mathrm{O}
$
To equalize the number of electrons equation (1) $\times 5$ and equation (2) $\times 2$
$
\begin{aligned}
& 5 \mathrm{SnCl}_2 \rightarrow 5 \mathrm{SnCl}_4+10 \mathrm{e}^{-} \\
& 2 \mathrm{KMnO}_4+16 \mathrm{HCl}+10 \mathrm{e}^{-} \rightarrow 2 \mathrm{MnCl}_2+4 \mathrm{H}_2 \mathrm{O}+2 \mathrm{KCl} \\
& 2 \mathrm{KMnO}_4+5 \mathrm{SnCl}_2+16 \mathrm{HCl} \rightarrow 5 \mathrm{SnCl}_4+2 \mathrm{MnCl}_2+4 \mathrm{H}_2 \mathrm{O}+2 \mathrm{KCl}
\end{aligned}
$
2. $\mathrm{C}_2 \mathrm{O}_4{ }^{2-}+\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} \rightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_2$ (in acid medium)
Oxidation half reaction:

Reduction half reaction:

To balance oxygen atoms, $\mathrm{H}_2 \mathrm{O}$ is added on RHS of equation (2)
$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$
To balance Hydrogen atoms, $\mathrm{H}^{+}$is added on LHS of equation (1) $\mathrm{C}_2 \mathrm{O}_4{ }^{2-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{CO}_2+2 \mathrm{e}^{-}$
To equalize the number of electrons gained and lost, multiply the equation (4) $\times 3$.

3. $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+\mathrm{NaI}$ (in acid medium)
Oxidation half reaction: (Loss of electron)
$
\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{e}^{2-}
$
Reduction half reaction: (Gain of electron)
$
\mathrm{I}_2+2 \mathrm{e}^{2-} \rightarrow 2 \mathrm{NaI}
$
Adding (1) and (2)

To balance oxygen, $2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_2+2 \mathrm{NaI}$
In acidic medium 4. $\mathrm{Zn}+\mathrm{NO}_3^{-} \rightarrow \mathrm{Zn}^{2+}+\mathrm{NO}$ Half reactions are -

TextQuestions - Evaluate Yourself
Question 1.
By applying the knowledge of chemical classification, classify each of the following Into elements, compounds or mixtures.
1. Sugar
2. Sea water
3. Distilled water
4. Carbon dioxide
5. Copper wire
6. Table salt
7. Silver plate
8. Naphthalene balls
Answer:

Question 2.
Calculate the molar mass of the following.
1. Ethanol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)$
2. Potassium permanganate $\left(\mathrm{KMnO}_4\right)$
3. Potassium dichromate $\left(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\right)$
4. Sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$
Answer:
(1) Ethanol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)$
Molar mass $=(2 \times 12)+(6 \times 1)+(1 \times 16)$ $=24+6+16=46 \mathrm{~g} \mathrm{~mol}^{-1}$
(2) Potassium permanganate $\left(\mathrm{KMnO}_4\right)$
Molar mass $=39+55+(4 \times 16)$ $=39+55+64=158 \mathrm{~g} \mathrm{~mol}^{-1}$
(3) Potassium dichromate $\left(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\right)$
Molar mass $=(39 \times 2)+(2 \times 52)+(7 \times 16)$ $=78+104+112=294 \mathrm{~g} \mathrm{~mol}^{-1}$
(4) Sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$
Molar mass $=(12 \times 12)+(22 \times 1)+(11 \times 16)=144+22+176=342 \mathrm{~g} \mathrm{~mol}^{-1}$
Question 3.
(a) Calculate the number of moles present in $9 \mathrm{~g}$ of ethane.
Answer:
Mass of ethane $=9 \mathrm{~g}$
Molar mass of ethane $\mathrm{C}_2 \mathrm{H}_6=30 \mathrm{~g} \mathrm{~mol}^{-1}$
No. of moles $=\frac{\text { Mass }}{\text { Molarmass }}=\frac{9}{30}=0.3 \mathrm{~mol}$.
(b) Calculate the number of molecules of oxygen gas that occupies a volume of $224 \mathrm{ml}$ at $273 \mathrm{~K}$ and $3 \mathrm{~atm}$ pressure.

Answer:
Molar volume of oxygen $=22400 \mathrm{ml}$.
$22400 \mathrm{ml}$ of oxygen contains $6.023 \times 10^{23}$ molecules.
$224 \mathrm{ml}$ of oxygen contain $\frac{6.023 \times 10^{23}}{22400} \times 224$
$\frac{6.023 \times 10^{23}}{100}=6.023 \times 10^{21}$
Question 4.
(a) $0.456 \mathrm{~g}$ of a metal gives $0.606 \mathrm{~g}$ of its chloride. Calculate the equivalent mass of the metal.
Answer:
Mass of the metal $=\mathrm{W}_1=0.456 \mathrm{~g}$.
Mass of the metal chloride $=\mathrm{W}_2=0.606 \mathrm{~g}$.
$\therefore$ Mass of chlorine $=\mathrm{W}_2-\mathrm{W}_1=0.606-0.456=0.15 \mathrm{~g}$
$0.15 \mathrm{~g}$ of chlorine combine with $0.456 \mathrm{~g}$ of metal.
$\therefore 35.46 \mathrm{~g}$ of chlorine will combine with $\frac{0.456}{0.15} \times 35.46=107.79 \mathrm{~g} \mathrm{eq}^{-1}$
(b) Calculate the equivalent mass of potassium dichromate. The reduction half-reaction in acid medium is -
$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$
Answer:
$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$
Equivalent mass $=\frac{\text { Molar mass }}{\text { No. of electrons gained }}$
Molar mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=294.18$
$\therefore$ Equivalent mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=\frac{294.18}{6}=49.03$
Question 5.
A Compound on analysis gave the following percentage composition $\mathrm{C}=54.55 \%, \mathrm{H}=9.09 \%, \mathrm{O}=$ $36.36 \%$. Determine the empirical formula of the compound.
Answer:

Question 6.
Experimental analysis of a compound containing the elements $x, y, z$ on analysis gave the following data, $x$ $=32 \%, \mathrm{y}=24 \%, \mathrm{z}=44 \%$. The relative number of atoms of $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are 2,1 and 0.5 , respectively.
(Molecular mass of the compound is $400 \mathrm{~g}$ ) Find out.
1. The atomic masses of the element $x, y, z$.
2. Empirical formula of the compound and
3. Molecular formula of the compound.
Answer:
Element $\mathrm{x}=32 \% ; \mathrm{y}=24 \% ; \mathrm{z}=44 \%$
Relative number of atoms $\mathrm{x}=2 ; \mathrm{y}=1 ; \mathrm{z}=0.5$
Molar mass of the compound $=400 \mathrm{~g}$.
1. Atomic mass of the element.
$
=\frac{\text { Percentage composition }}{\text { Atomic mass }}
$
Relative number of atoms Atomic mass
$
=\frac{\text { Percentage composition }}{\text { Relative No. of atoms }}
$
$\therefore$ Atomic mass Relative Atomic mass $\mathrm{x}=\frac{32}{2}=16$
Atomic mass $\mathrm{y}=\frac{24}{1}=24$
Atomic mass $\mathrm{z}=\frac{44}{0.5}=88$

2. Empirical formula of the compound $x_4 y_2 Z_1$
Molecular mass of the compound $=400$
$
n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{400}{200}=2
$
3. Molecular formula of the compound $=x_8 y_4 z_2$
Question 7.
The balanced equation for a reaction is given below $2 x+3 y \rightarrow 41+\mathrm{m}$ When 8 moles of $\mathrm{x}$ react with 15 moles of $y$, then -
1. Which is the limiting reagent?
2. Calculate the amount of products formed.
3. Calculate the amount of excess reactant left at the end of the reaction.
Answer:
$
2 \mathrm{x}+3 \mathrm{y} \rightarrow 41+\mathrm{m}
$
1. $2 \mathrm{x}$ reacts with $3 \mathrm{y}$ to give products.
$5 \mathrm{x}$ reacts with $15 \mathrm{y}$ means, $\mathrm{y}$ is the excess because 8 moles of $\mathrm{x}$ should react withn $4 \mathrm{x} 3 \mathrm{y}=12 \mathrm{y}$ moles of $\mathrm{y}$ to give products. In this reaction $15 \mathrm{y}$ moles are used. Therefore, 3 moles of $y$ is excess and it is the limiting agent.
2. When 8 moles of $x$ react with 12 moles of $y$, the product formed will be $4 \times 41$ i.e. 161 and $4 \mathrm{~m}$ as product.
$
8 \mathrm{x}+12 \mathrm{y} \rightarrow 161+4 \mathrm{~m}
$
3. At the end of the reaction, the excess reactant left in 3 moles of $y$.
Question 8.
Balance the following equation using oxidation number method
$
\mathrm{AS}_2 \mathrm{~S}_3+\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{ASO}_4+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{NO}
$
Answer:

To balance oxygen and sulphur, $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{SO}_4$ are added.
$
\begin{aligned}
& 3 \mathrm{AS}_2 \mathrm{~S}_3+2 \mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow 6 \mathrm{H}_3 \mathrm{ASO}_4+2 \mathrm{NO}+\mathrm{H}_2 \mathrm{SO}_2 \\
& 3 \mathrm{AS}_2 \mathrm{~S}_3+28 \mathrm{HNO}_3+4 \mathrm{H}_2 \mathrm{O} \rightarrow 6 \mathrm{H}_3 \mathrm{ASO}_4+28 \mathrm{NO}+9 \mathrm{H}_2 \mathrm{SO}_4
\end{aligned}
$
Textual Calculations based on Stolchlometry Solved
Question 1.
How many moles of hydrogen is required to produce 10 moles of ammonia?
Answer:
The balanced stoichiometric equation fòr the formation of ammonia is
$
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_2(\mathrm{~g})
$
4 s per the stoichiometric equation, to produce 2 moles of ammonia, 3 moles of hydrogen are required.
$\therefore$ to produce 10 moles of ammonia,

$=15$ moles of hydrogen are required.
Question 2.
Calculate the amount of water produced by the combustion of $32 \mathrm{~g}$ of methane.
Answer:
$
\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}
$
As per the stoichiometric equation,
Combustion of 1 mole $(16 \mathrm{~g}) \mathrm{CH}_4$ produces 2 moles $(2 \times 18=36 \mathrm{~g})$ of water.

Question 3.
How much volume of carbon dioxide is produced when $50 \mathrm{~g}$ of calcium carbonate is heated completely under standard conditions?
Answer:
The balanced chemical equation is,
$
\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})
$
As per the stoichiometric equation, 1 mole $(100 \mathrm{~g}) \mathrm{CaCO}_3$ on heating produces 1 mole $\mathrm{CO}_2$

$=11.35$ litres of $\mathrm{CO}_2$
Question 4.
How much volume of chlorine is required to form $11.2 \mathrm{~L}$ of $\mathrm{HCl}$ at $273 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure?
Answer:
The balanced equation for the formation of $\mathrm{HCl}$ is,
$
\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})
$
As per the stoichiometric equation, under given conditions, To produce 2 moles of $\mathrm{HCl}, 1$ mole of chlorine gas is required.
To produce 44.8 liters of $\mathrm{HCl}, 22.4$ liters of chlorine gas are required
$\therefore$ To produce 11.2 liters of $\mathrm{HCl}$,

$=5.6$ litres of chlorine are required.
Question 5.
Calculate the percentage composition of the elements present in magnesium carbonate. How many kilogram of $\mathrm{CO}_2$ can be obtained by heating $1 \mathrm{~kg}$ of $90 \%$ pure magnesium carbonate.
Answer:
The balanced chemical equation is
$
\mathrm{MgCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+\mathrm{CO}_2
$
Molar mass of $\mathrm{MgCO}_3$ is $84 \mathrm{~g} \mathrm{~mol}^{-1}$.
$84 \mathrm{~g} \mathrm{MgCO}_3$ contain $24 \mathrm{~g}$ of Magnesium.
$\therefore 100 \mathrm{MgCO}_3$ contain

$\begin{aligned}
& =28.57 \mathrm{~g} \mathrm{Mg} \text { i.e. percentage of magnesium }=28.57 \% . \\
& 84 \mathrm{~g} \mathrm{MgCO}_3 \text { contain } 12 \mathrm{~g} \text { of carbon }
\end{aligned}$

$=14.29 \mathrm{~g}$ of carbon
Percentage of carbon $=14.29 \%$.
$84 \mathrm{~g} \mathrm{MgCO}_3$ contain $48 \mathrm{~g}$ of oxygen
$\therefore 100 \mathrm{~g} \mathrm{MgCO}_3$ contains

$=57.14 \mathrm{~g}$ of oxygen.
$\therefore$ Percentage of oxygen $=57.14 \%$.
As per the stoichiometric equation,
$84 \mathrm{~g}$ of $100 \%$ pure $\mathrm{Mg} \mathrm{CO}_3$ on heating gives $44 \mathrm{~g}$ of $\mathrm{CO}_2$.
$\therefore 1000$ g of $90 \%$ pure $\mathrm{Mg} \mathrm{CO}_3$ gives
$
\begin{aligned}
& \frac{44 g}{84 g x 100} \times 90 \% \times 1000 \mathrm{~g} \\
& =471.43 \mathrm{~g} \text { of } \mathrm{CO}_2 \\
& =0.471 \mathrm{~kg} \text { of } \mathrm{CO}_2
\end{aligned}
$
Question 6.

(1) If the entire quantity of all the reactants is not consumed in the reaction which is the limiting reagent?
(2) Calculate the quantity of urea formed and un reacted quantity of the excess reagent. The balanced equation is

Answer:
(1) The entire quantity of ammonia is consumed in the reaction. So ammonia is the limiting reagent. Some quantity of $\mathrm{CO}_2$ remains unreacted, so $\mathrm{CO}_2$ is the excess reagent.
(2) Quantity of urea formed = number of moles of urea formed $x$ molar mass of urea $=19$ moles $\mathrm{x} 60 \mathrm{~g} \mathrm{~mol}^{-1}$ $=1140 \mathrm{~g}=1.14 \mathrm{~kg}$
Excess reagent leftover at the end of the reaction is carbon dioxide. Amount of carbon dioxide leftover $=$ number of moles of $\mathrm{CO}_2$ left over =7 \text { $moles } \mathrm{x} 44 \mathrm{~g} \mathrm{~mol}^{-1}=308 \mathrm{~g} \text {. }$