Text Book Back Questions and Answers - Chapter 9 Solutions 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Solutions
Textual Evaluation Solved
Multiple Choice Questions
Question 1.
The molality of a solution containing $1.8 \mathrm{~g}$ of glucose dissolved in $250 \mathrm{~g}$ of water is
(a) $0.2 \mathrm{M}$
(b) $0.01 \mathrm{M}$
(c) $0.02 \mathrm{M}$
(d) $0.04 \mathrm{M}$
Answer:
(d) $0.04 \mathrm{M}$
Solution:
$
\text { Molality }=\frac{\text { Number of moles of solute }}{\text { Weight of solvent (in kg) }}=\frac{\left(\frac{1.8}{180}\right)}{0.25}=\frac{0.01}{0.25}=0.04 \mathrm{M}
$
Question 2.
Which of the following concentration terms is/are independent of temperature?
(a) molality
(b) molarity
(c) mole fraction
(d) (a) and (c)
Answer:
(d) (a) and (c)
Solution:
Molality and mole fraction are independent of temperature.
Question 3.
Stomach acid, a dilute solution of $\mathrm{HCI}$ can be neutralised by reaction with Aluminium hydroxide $\mathrm{Al}(\mathrm{OH})_3+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_3+3 \mathrm{H}_2 \mathrm{O}$
How many millilitres of $0.1 \mathrm{MAl}(\mathrm{OH})_3$ solution are needed to neutralise $21 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$
(a) $14 \mathrm{~mL}$
(b) $7 \mathrm{~mL}$
(c) $21 \mathrm{~mL}$
(d) none of these
Answer:
(b) $7 \mathrm{~mL}$
Solution:
$\mathrm{M}$
$
1 \times \mathrm{V}_1=\mathrm{M}_2 \times \mathrm{V}_2
$
$\because 0.1 \mathrm{M} \mathrm{Al}(\mathrm{OH})_3$ gives $3 \times 0.1=0.3 \mathrm{M} \mathrm{OH}^{-}$ions .

$
\begin{aligned}
& 0.3 \times \mathrm{V}_1=0.1 \times 21 \\
& \mathrm{~V}_1=\frac{0.1 \times 21}{0.3}=7 \mathrm{ml}
\end{aligned}
$
Question 4.
The partial pressure of nitrogen in air is $0.76 \mathrm{~atm}$ and its Henry's law constant is $7.6 \times 10^4$ atm at $300 \mathrm{~K}$. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at $300 \mathrm{~K}$ ?
(a) $1 \times 10^{-4}$
(b) $1 \times 10^{-6}$
(c) $2 \times 10^{-5}$
(d) $1 \times 10^{-5}$
Answer:
(d) $1 \times 10^{-5}$
Solution:
$
\begin{aligned}
& \mathrm{P}_{\mathrm{N}_2}=0.76 \mathrm{~atm} \\
& \mathrm{~K}_{\mathrm{H}}=7.6 \times 10^4 \\
& \mathrm{x}=? \\
& \mathrm{P}_{\mathrm{N}_2}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{x} \\
& 0.76=7.6 \times 10^4 \times \mathrm{x} \\
& \mathrm{x}=\frac{0.76}{7.6 \times 10^4}=1 \times 10^{-5}
\end{aligned}
$
Question 5 .
The Henry's law constant for the solubility of Nitrogen gas in water at $350 \mathrm{~K}$ is $8 \times 10^4 \mathrm{~atm}$. The mole fraction of nitrogen in air is 0.5 . The number of moles of Nitrogen from air dissolved in 10 moles of water at $350 \mathrm{~K}$ and $4 \mathrm{~atm}$ pressure is
(a) $4 \times 10^{-4}$
(b) $4 \times 10^4$
(c) $2 \times 10^{-2}$
(d) $2.5 \times 10^{-4}$
Answer:
(d) $2.5 \times 10^{-4}$
Solution:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{H}}=8 \times 10^4 \\
& \left(\mathrm{x}_{\mathrm{N}_2}\right)_{\text {in air }}=0.5
\end{aligned}
$
Total pressure $=4 \mathrm{~atm}$
Partial pressure of nitrogen $=$ Mole fraction Total pressure

$
\begin{aligned}
& =0.5 \times 4=2 \\
& \left(\mathrm{P}_{\mathrm{N}_2}\right)=\mathrm{K}_{\mathrm{H}} \times \text { Mole fraction of } \mathrm{N}_2 \text { in solution } \\
& 2=8 \times 10^4 \times \frac{\text { Number of moles of nitrogen }}{\text { Total number of moles }} \\
& \frac{10+\text { No. of moles of } \mathrm{N}_2}{\text { No. of moles of } \mathrm{N}_2}=\frac{8 \times 10^4}{2} \\
& \frac{10}{\text { No. of moles of } \mathrm{N}_2}+1=4 \times 10^4 \\
& \frac{10}{\text { No. of moles of } \mathrm{N}_2}=40000-1 \\
& \text { No. of moles of } \mathrm{N}_2=\frac{10}{39999}=2.5 \times 10^{-4}
\end{aligned}
$
Question 6.
Which one of the following is incorrect for ideal solution?
(a) $\Delta \mathrm{H}_{\text {mix }}=0$
(b) $\Delta \mathrm{U}_{\text {mix }}=0$
(c) $\Delta \mathrm{P}=\mathrm{P}_{\mathrm{Observed}}-\mathrm{P}_{\mathrm{Calculated} \text { by raoults law }}=0$
(d) $\Delta \mathrm{G}_{\text {mix }}=0$
Answer:
(d) $\Delta \mathrm{G}_{\text {mix }}=0$
Solution:
For an ideal solution, $\Delta \mathrm{S}_{\text {mix }} \neq 0$; Hence $\Delta \mathrm{G}_{\text {mix }} \neq 0$
$\therefore$ Incorrect is $\Delta \mathrm{G}_{\text {mix }}=0$
Question 7.
Which one of the following gases has the lowest value of Henry's law constant?
(a) $\mathrm{N}_2$
(b) $\mathrm{He}$
(c) $\mathrm{CO}_2$
(d) $\mathrm{H}_2$
Answer:
(c) $\mathrm{CO}_2$
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry's Law constant.

Question 8.
$\mathrm{P}_1$ and $\mathrm{P}_2$ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if $\mathrm{x}_1$ represents the mole fraction of component 1 , the total pressure of the solution formed by 1 and 2 will be
(a) $P_1+x_1\left(P_2-P_1\right)$
(b) $\mathrm{P}_2-\mathrm{x}_1\left(\mathrm{P}_2+\mathrm{P}_1\right)$
(c) $\mathrm{P}_1-\mathrm{x}_2\left(\mathrm{P}_1-\mathrm{P}_2\right)$
(d) $P_1+x_2\left(P_1-P_2\right)$
Answer:
(c) $P_1-x_2\left(P_1-P_2\right)$
Solution:
$
\begin{aligned}
& \mathrm{P}_{\text {total }}=\mathrm{P}_1+\mathrm{P}_2 \\
& =\mathrm{P}_1 \mathrm{x}_1+\mathrm{P}_2 \mathrm{x}_2 \\
& =\mathrm{P}_1\left(1-\mathrm{x}_2\right)+\mathrm{P}_2 \mathrm{x}_2 \\
& =\mathrm{P}_1-\mathrm{P}_1 \mathrm{x}_2+\mathrm{P}_2 \mathrm{x}_2=\mathrm{P}_1-\mathrm{x}_2\left(\mathrm{P}_1-\mathrm{P}_2\right) \\
& {\left[\because \mathrm{x}_1+\mathrm{x}_2=1\right.} \\
& \left.\mathrm{x}_1=1-\mathrm{x}_2\right]
\end{aligned}
$
Question 9.
Osomotic pressure $(\pi)$ of a solution is given by the relation
(a) $\pi=n R T$
(b) $\pi \mathrm{V}=\mathrm{nRT}$
(c) $\pi \mathrm{RT}=\mathrm{n}$
(d) none of these
Answer:
(b) $\pi \mathrm{V}=\mathrm{nRT}$
Solution:
$
\begin{aligned}
& \mathrm{n}=\mathrm{CRT} \\
& \mathrm{n}=\frac{n}{V} \\
& \pi \mathrm{V}=\mathrm{nRT}
\end{aligned}
$

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
(a) Acetone + chloroform
(b) Water + nitric acid
(c) $\mathrm{HCI}+$ water
(d) ethanol + water
Answer:
(d) ethanol + water
Question 11.
The Henry's law constants for two gases $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{x}$ and $\mathrm{y}$ respectively. The ratio of mole fractions of $A$ to $B$ is 0.2 . The ratio of mole fraction of $B$ and $A$ dissolved in water will be
(a) $\frac{2 x}{y}$
(b) $\frac{y}{0.2 x}$
(c) $\frac{0.2 x}{y}$
(d) $\frac{5 x}{y}$
Answer:
(d) $\frac{5 x}{y}$
Solution:

Given,
$
\begin{gathered}
\left(\mathrm{K}_{\mathrm{H}}\right)_{\mathrm{A}}=x,\left(\mathrm{~K}_{\mathrm{H}}\right)_{\mathrm{B}}=y, \frac{x_{\mathrm{A}}}{x_{\mathrm{B}}}=0.2,\left(\frac{x_{\mathrm{B}}}{x_{\mathrm{A}}}\right)_{\text {in solution }}=? \\
\mathrm{P}_{\mathrm{A}}=x\left(x_{\mathrm{A}}\right)_{\text {in solution }} \\
\mathrm{P}_{\mathrm{B}}=y\left(x_{\mathrm{B}}\right)_{\text {in solution }} \\
\left(\frac{x_{\mathrm{B}}}{x_{\mathrm{A}}}\right)_{\text {in solution }}=\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}} \times \frac{x}{y}=\frac{x_{\mathrm{B}}}{x_{\mathrm{A}}} \times \frac{x}{y}=\frac{1}{0.2} \times \frac{x}{y}=\frac{5 x}{y}
\end{gathered}
$
Question 12.
At $100^{\circ} \mathrm{C}$ the vapour pressure of a solution containing $6.5 \mathrm{~g}$ a solute in $100 \mathrm{~g}$ water is $732 \mathrm{~mm}$. If $\mathrm{K}_{\mathrm{b}}=0.52$, the boiling point of this solution will be
(a) $102^{\circ} \mathrm{C}$
(b) $100^{\circ} \mathrm{C}$
(c) $101^{\circ} \mathrm{C}$
(d) $100.52^{\circ} \mathrm{C}$
Answer:
(c) $101^{\circ} \mathrm{C}$
Solution:
$
\begin{aligned}
\frac{\Delta P}{P^{\circ}} & =\frac{n_2}{n_1} \\
\mathrm{~W}_2 & =6.5 \mathrm{~g} \\
\mathrm{~W}_1 & =100 \mathrm{~g} \\
\mathrm{~K}_{\mathrm{b}} & =0.52 \\
\frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}} & =\frac{\mathrm{W}_2 \mathrm{M}_1}{\mathrm{M}_2 \mathrm{~W}_1} \\
\frac{760-732}{760} & =\frac{6.5 \times 18}{\mathrm{M}_2 \times 100} \\
\mathrm{M}_2 & =31.75 \\
\Delta \mathrm{T}_b & =\mathrm{K}_b \cdot m=\frac{0.52 \times 6.5 \times 1000}{31.75 \times 100}=1.06 \\
\mathrm{~T}_b-100 & =1.06 \\
\mathrm{~T}_b & =100+1.06=101.06 \approx 101^{\circ} \mathrm{C}
\end{aligned}
$

Question 13.
According to Raoults law, the relative Lowering of vapour pressure for a solution is equal to....
(a) molefraction of solvent
(b) mole fraction of solute
(c) number of moles of solute
(d) number of moles of solvent
Answer:
(b) mole fraction of solute
Solution:
$\frac{\Delta P}{P^*}=\mathrm{x}_2$ (Mole fraction of the solute)
Question 14.
At same temperature. which pair of the following solutions are isotonic?
(a) $0.2 \mathrm{M} \mathrm{BaCl}_2$ and $0.2 \mathrm{M}$ urea
(b) $0.1 \mathrm{M}$ glucose and $0.2 \mathrm{M}$ urea
(c) $0.1 \mathrm{MNaCl}$ and $0.1 \mathrm{MK}_2 \mathrm{SO}_4$
(d) $0.1 \mathrm{MBa}\left(\mathrm{NO}_3\right)_2$ and $0.1 \mathrm{MNa}_2 \mathrm{SO}_4$
Answer:
(d) $0.1 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_3\right)_2$ and $0.1 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_4$
Solution:
$0.1 \times 3$ ion $\left[\mathrm{Ba}^2+2 \mathrm{NO}_3{ }^{-}\right], 0.1 \times 3$ ion $\left[2 \mathrm{Na}^{+}, \mathrm{SO}_4{ }^{-}\right]$
Question 15.
The empirical formula of a non-electrolyte( $\mathrm{X})$ is $\mathrm{CH}_2 \mathrm{O}$. A solution containing six gram of $\mathrm{X}$ exerts the same osmotic pressure as that of $0.025 \mathrm{M}$ glucose solution at the same temperature. The molecular formula of $\mathrm{X}$ is
(a) $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2$
(b) $\mathrm{C}_8 \mathrm{H}_{16} \mathrm{O}_8$
(c) $\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_4$
(d) $\mathrm{CH}_2 \mathrm{O}$
Answer:
(b) $\mathrm{C}_8 \mathrm{H}_{16} \mathrm{O}_8$
Solution:

$\left(\pi_1\right)_{\text {non electrolute }}=\left(\pi_2\right)_{\text {glucose }}$
$
\begin{aligned}
& \mathrm{C}_1 \mathrm{RT}=\mathrm{C}_2 \mathrm{RT} \\
& \frac{\mathrm{W}_1}{\mathrm{M}_1}=\frac{\mathrm{W}_2}{\mathrm{M}_2} \quad 12+2+16=30 \\
& \frac{6}{n(30)}=0.025 \\
& \mathrm{n}=\frac{6}{0.025 x 30}=30 \\
& \therefore \text { Molecular formula } \mathrm{CH}_8 \mathrm{H}_{16} \mathrm{O}_8
\end{aligned}
$
Question 16.
The $\mathrm{K}_{\mathrm{H}}$ for the solution of oxygen dissolved in water is $4 \times 10^4$ atm at a given temperature. If the partial pressure of oxygen in air is $0.4 \mathrm{~atm}$, the mole fraction of oxygen in solution is
(a) $4.6 \times 10^3$
(b) $1.6 \times 10^4$
(c) $1 \times 10^{-5}$
(d) $1 \times 10^5$
Answer:
(c) $1 \times 10^{-5}$
Solution:

$
\begin{aligned}
& \mathrm{K}_{\mathrm{H}}=4 \times 10^4 \mathrm{~atm} \\
& \left(\mathrm{P}_{\mathrm{O}_2}\right)_{\text {air }}=0.4 \mathrm{~atm} \\
& \left(\mathrm{x}_{\mathrm{O}_2}\right)_{\text {in solution }}=? \\
& \text { air }- \text { in solution } \\
& \left(\mathrm{P}_{\mathrm{O}_2}\right)_{\text {air }}=\mathrm{K}_{\mathrm{H}}\left(\mathrm{x}_{\mathrm{o}_2}\right)_{\text {in solution }} \\
& 0.4=4 \times 10^4\left(\mathrm{x}_{\mathrm{O}_2}\right)_{\text {in solution }} \\
& \left(\mathrm{x}_{\mathrm{O}_2}\right)_{\text {in solution }}=\frac{0.4}{4 \times 10^4}=1 \times 10^{-5}
\end{aligned}
$
Question 17.
Normality of $1.25 \mathrm{M}$ sulphuric acid is
(a) $1.25 \mathrm{~N}$
(b) $3.75 \mathrm{~N}$
(c) $2.5 \mathrm{~N}$
(d) $2.25 \mathrm{~N}$
Answer:
(c) $2.5 \mathrm{~N}$
Solution:
Normality of $\mathrm{H}_2 \mathrm{SO}_4=\left(\mathrm{No}\right.$. of replacable $\left.\mathrm{H}^{+}\right) \times \mathrm{M}=2 \times 1.25=2.5 \mathrm{~N}$
Question 18.
Two liquids $\mathrm{X}$ and $\mathrm{Y}$ on mixing gives a warm solution. The solution is
(a) ideal
(b) non-ideal and shows positive deviation from Raoults law
(c) ideal and shows negative deviation from Raoults Law
(d) non - ideal and shows negative deviation from Raoults Law
Answer:
(d) non - ideal and shows negative deviation from Raoults Law
Solution:
$\Delta \mathrm{H}_{\operatorname{mix}}$ is negative and show negative deviation from Raoults law.
Question 19.
The relative lowering of vapour pressure of a sugar solution in water is $3.5 \times 10^{-3}$. The mole fraction of water in that solution is
(a) 0.0035
(b) 0.35
(c) $0.0035 / 18$
(d) 0.9965
Answer:
(d) 0.9965

Solution:
$
\begin{aligned}
\frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}} & =x_{\text {sugar }} \\
3.5 \times 10^{-3} & =\mathrm{x}_{\text {sugar }} \\
x_{\text {sugar }}+x_{\mathrm{H}_2 \mathrm{O}} & =1 \\
x_{\mathrm{H}_2 \mathrm{O}} & =1-0.0035=0.9965
\end{aligned}
$
Question 20.
The mass of a non-volatile solute (molar mass $80 \mathrm{~g} \mathrm{~mol}^{-1}$ ) which should be dissolved in $92 \mathrm{~g}$ of toluene to reduce its vapour pressure to $90 \%$.........
(a) $10 \mathrm{~g}$
(b) $20 \mathrm{~g}$
(c) $9.2 \mathrm{~g}$
(d) $8.89 \mathrm{~g}$
Answer:
(d) $8.89 \mathrm{~g}$
Solution:
$
\begin{aligned}
& \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=x_2 \\
& \frac{100-90}{100}=\frac{n_2}{n_2+n_1} \Rightarrow \frac{n_2+n_1}{n_2}=\frac{100}{10} \quad\left[n_1=\frac{92}{92}=1\right] \\
& 1+\frac{1}{n_2}=10 \\
& \frac{1}{n_2}=9 \Rightarrow n_2=\frac{1}{9} \\
& \frac{W_2}{M_2}=\frac{1}{9} \Rightarrow W_2=\frac{M_2}{9} \\
& \mathrm{~W}_2=\frac{80}{9}=8.89 \mathrm{~g} \\
&
\end{aligned}
$
Question 21.
For a solution, the plot of osmotic pressure $(\pi)$ verses the concentration $\left(e\right.$ in $\left.\mathrm{mol} \mathrm{L}^{-1}\right)$ gives a straight line with slope $310 \mathrm{R}$ where ' $\mathrm{R}$ ' is the gas constant. The temperature at which osmotic pressure measured is
(a) $310 \times 0.082 \mathrm{~K}$
(b) $310^{\circ} \mathrm{C}$

(c) $37^{\circ} \mathrm{C}$
(d) $\frac{310}{20.082}$
Answer:
(c) $37^{\circ} \mathrm{C}$
Solution:
$
\begin{aligned}
& \pi=\mathrm{CRT} \\
& \mathrm{y}=\mathrm{x}(\mathrm{m}) \\
& \mathrm{m}=\mathrm{RT} \\
& 310 \mathrm{R}=\mathrm{RT} \\
& \mathrm{T}=310 \mathrm{~K} \\
& =37^{\circ} \mathrm{C}
\end{aligned}
$
Question 22.
$200 \mathrm{ml}$ of an aqueous solution of a protein contains $1.26 \mathrm{~g}$ of protein. At $300 \mathrm{~K}$, the osmotic pressure of this solution is found to be $2.52 \times 10^{-3}$ bar. The molar mass of protein will be $\left(\mathrm{R}=0.083 \mathrm{Lhar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$
(a) $62.22 \mathrm{Kg} \mathrm{mol}^{-1}$
(b) $12444 \mathrm{~g} \mathrm{~mol}^{-1}$
(c) $300 \mathrm{~g} \mathrm{~mol}^{-1}$
(d) none of these
Answer:
(a) $62.22 \mathrm{Kg} \mathrm{mol}^{-1}$
Solution:
$
\begin{aligned}
& \pi=\mathrm{CRT} \\
& \mathrm{M}=\frac{W R T}{\pi 1}=\frac{1.26 \times 0.083 \times 300}{2.52 \times 10^{-3} \times 0.2}=62.22 \mathrm{Kg} \mathrm{mol}^{-1}
\end{aligned}
$
Question 23.
The Van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:

(b) 1
Solution:
$\mathrm{Ba}(\mathrm{OH})_2$ dissociates to form $\mathrm{Ba}^{2+}$ and $2 \mathrm{OH}^{-1}$ ion
$
\begin{aligned}
& \alpha=\frac{(i-1)}{(n-1)} \\
& \mathrm{i}=\alpha(\mathrm{n}-1)+1 \\
& \mathrm{n}=\mathrm{i}=3\left(\text { for } \mathrm{Ba}(\mathrm{OH})_2, \alpha=1\right)
\end{aligned}
$
Question 24 .
What is the molality of a $10 \% \mathrm{w} / \mathrm{w}$ aqueous sodium hydroxide solution?
(a) 2.778
(b) 2.5
(c) 10
(a) 0.4
Answer:
(b) 2.5
Solution:
$100 \% \frac{w}{w}$ aqueous $\mathrm{NaOH}$ solution means that $10 \mathrm{~g}$ of sodium hydroxide in $100 \mathrm{~g}$ solution.
$
\text { Molality }=\frac{\text { No. of moles of solute }}{\text { Weight of solvent (in kg) }}=\frac{\left(\frac{10}{40}\right)}{0.1}=\frac{0.25}{0.1}=2.5 \mathrm{M}
$
Question 25 .
The correct equation for the degree of an associating solute, ' $n$ ' molecules of which undergoes association in solution, is
(a) $\alpha=\frac{n(i-1)}{n-1}$
(b) $\alpha^2=\frac{n(1-i)}{n-1}$
(c) $\alpha=\frac{n(i-1)}{1-n}$
(d) $\alpha=\frac{n(1-i)}{n(1-i)}$

Answer:
(c) $\alpha=\frac{n(i-1)}{1-n}$
Solution:
$\alpha=\frac{(i-1) n}{(n-1)}$ (or) $\frac{n(i-1)}{(1-n)}$
Question 26.
Which of the following aqueous solutions has the highest boiling point?
(a) $0.1 \mathrm{M} \mathrm{KNO}_3$
(b) $0.1 \mathrm{M} \mathrm{Na}_3 \mathrm{PO}_4$
(c) $0.1 \mathrm{M} \mathrm{BaCl}_2$
(d) $0.1 \mathrm{M} \mathrm{K}_2 \mathrm{SO}_4$
Answer:
(a) $0.1 \mathrm{M} \mathrm{KNO}_3$
Solution:
Elevation of boiling point is more in the case of $\mathrm{Na}_3 \mathrm{PO}_4$ (no. of ions $4 ; 3 \mathrm{Na}^{+}, \mathrm{PO}_4{ }^{3-}$ )
Question 27.
The freezing point depression constant for water is $1.86^{\circ} \mathrm{k} \mathrm{kg} \mathrm{mol}{ }^{-1}$. If $5 \mathrm{~g} \mathrm{Na}_2 \mathrm{SO}_4$ is dissolved in $45 \mathrm{~g}$ water, the depression in freezing point is $3.64^{\circ} \mathrm{C}$. The van't Hoff factor for $\mathrm{Na}_2 \mathrm{SO}_4$ is .......
(a) 2.50
(b) 2.63
(c) 3.64
(d) 5.50
Answer:
(a) 2.50
Solution:
$\mathrm{K}_{\mathrm{f}}=1.86$
$\mathrm{W}_2=5 \mathrm{~g}$
$\Delta \mathrm{T}_{\mathrm{f}}=3.64$
$\mathrm{M}_2=142$
$\mathrm{w}_1=45 \mathrm{~g}$

$
\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ix} \mathrm{K}_{\mathrm{f}} \\
& i=\frac{\Delta \mathrm{T}_f \times \mathrm{M}_2 \times \mathrm{W}_1}{\mathrm{~K}_f \times \mathrm{W}_2 \times 1000} \\
& =\frac{3.64 \times 142 \times 45}{1.86 \times 5 \times 1000} \\
& i=2.5 \\
&
\end{aligned}
$
Question 28.
Equimolal aqueous solutions of $\mathrm{NaCI}$ and $\mathrm{KCI}$ are prepared. If the freezing point of $\mathrm{NaCI}$ is $-2^{\circ} \mathrm{C}$, the freezing point of $\mathrm{KCI}$ solution is expected to be
(a) $-2^{\circ} \mathrm{C}$
(b) $-4^{\circ} \mathrm{C}$
(c) $-1^{\circ} \mathrm{C}$
(d) $0^{\circ} \mathrm{C}$
Answer:
(a) $-2^{\circ} \mathrm{C}$
Solution:
Equimolal aqueous solution of $\mathrm{KCI}$ also shows $2^{\circ} \mathrm{C}$ depression in freezing point.
Question 29.
Phenol dimerises in henzene having van't Hoff factor 0.54 . What is the degree of association?
(a) 0.46
(b) 92
(c) 46
(d) 0.92
Answer:
(d) 0.92
Solution:
$
\alpha=\frac{(1-i) n}{(n-1)}=\frac{(1-0.54) 2}{(2-1)}=0.46 \times 2=0.92
$
Question 30.
Assertion : An ideal solution obeys Raoults Law
Reason : In an ideal solution, solvent-solvent as well as solute-solute interactions are similar to solutesolvent interactions.
(a) both assertion and reason are true and 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:
(a) both assertion and reason are true and reason is the correct explanation of assertion

Short Answer Questions
Question 31 .
Define
1. Molality
2. Normality
Answer:
1. Molality (m):
It is defined as the number of moles of the solute present in $1 \mathrm{~kg}$ of the solvent $m=\frac{\text { Number of moles of solute }}{\text { Mass of the solvent (in } \mathrm{kg} \text { ) }}$
2. Normality $(\mathrm{N})$ :
It is defined as the number of gram equivalents of solute in I litre of the solution. $N=\frac{\text { Number of gram equivalent of solute }}{\text { Volume of solution (in L) }}$
Question 32 .
What is a vapour pressure of liquid? What is relative lowering of vapour pressure?
Answer:
1. The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature.
2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour. pressure to vapour pressure of pure solvent. Relative lowering of vapour pressure
$
=\frac{p_{\text {solvent }}^{\circ}-p_{\text {solution }}}{p_{\text {solvent }}^{\circ}}
$

Question 33.
State and explain Henry's law.
Answer:
Henry's law:
This law states "that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations."
$\mathrm{P}_{\text {solute }} \propto \mathrm{x}_{\text {solute in solution }}$
$\mathrm{P}_{\text {solute }}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{x}_{\text {solute in solution }}$
$\mathrm{x}_{\text {solute }}=$ mole fraction of solute in the solution
$\mathrm{K}_{\mathrm{H}}=$ empirical constant.
$\mathrm{P}_{\text {solute }}=$ Vapour pressure of the solute (or) the partial pressure of the gas in vapour state. The value of $\mathrm{K}_{\mathrm{H}}$ depends on the nature of the gaseous solute and solvent.
Question 34.
State Raoult law and obtain expression for lowering of apour pressure when nonvolatile solute is dissolved in solvent.
Answer:
Raoult's law:
This law states that "in the case of a solution of volatile liquids the partial vapour pressure of each component (A \& B) of the solution is directly proportional to its mole fraction.
$
\mathrm{P}_{\mathrm{A}} \propto \mathrm{x}_{\mathrm{A}}
$

when $\mathrm{x}_{\mathrm{A}}=1$,
then $\mathrm{k}=\mathrm{P}^{\circ} \mathrm{A}$
( $\mathrm{P}^{\circ} \mathrm{A}=$ vapour pressure of pure component)
$\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}} \cdot \mathrm{xa}$
$\mathrm{P}_{\mathrm{B}}=\mathrm{P}^{\circ} \mathrm{B} \cdot \mathrm{xb}$
when a non volatile is dissolved in pure water, the vapour pressure of the pure solvent will decrease. In such solution, the vapours pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.
$\mathrm{P}_{\text {solution }} \propto \mathrm{x}_{\mathrm{A}}$
$P_{\text {solution }}=\mathrm{k} \cdot \mathrm{x}_{\mathrm{A}}$
$\mathrm{x}_{\mathrm{A}}=1, \mathrm{k}=\mathrm{P}_{\text {solvent }}^{\circ}$
$\mathrm{P}_{\text {solution }}^{\circ}=\mathrm{P}_{\text {solvent }}^{\circ}-\mathrm{P}_{\text {solution }}$
Lowering of vapour pressure $=\mathrm{P}^{\circ}$ solvent $-\mathrm{P}_{\text {solution }}$
Relative lowering of vapour pressure $=\frac{P^x-P}{P^{\prime}}=\mathrm{x}_{\mathrm{B}}$
where $\mathrm{x}_{\mathrm{B}}=$ Mole fraction of solute.
Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
$\mathrm{K}_{\mathrm{f}}=$ molar freezing point depression constant or cryoscopic constant.
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$,
where
$\Delta \mathrm{T}_{\mathrm{f}}=$ depression in freezing point.
$\mathrm{m}=$ molality of the solution
$\mathrm{K}_{\mathrm{f}}=$ cryoscopic constant
If $m=I$
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}}$
i.e., cryoscopic constant is equal to the depression in freezing point for 1 molal solution cryoscopic constant depends on the molar concentration of the solute particles. $\mathrm{K}_{\mathrm{f}}$ is directly proportional to the molal concentration of the solute particles.
$\Delta \mathrm{T}_f=\mathrm{K}_f \times \frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}$
$\mathrm{W}_{\mathrm{B}}=$ mass of the solute
$\mathrm{W}_{\mathrm{A}}=$ mass of solvent
$\mathrm{M}_{\mathrm{B}}=$ molecular mass of the solute.
Question 36.
What is osmosis?
Answer:
Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.

Question 37.
Define the term bisotonic
Answer:
1. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
2. When such solutions arc separated by a semipermeable membrane, solvent flow between one to the other on either direction is same. i.e.. the net solvent flow between two isotonic solutions is zero.
Long Answer Questions
Question 38 .
You are provided with a solid 'A' and three solutions of $\mathrm{A}$ dissolved in water - one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
1. Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.
2. Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.
3. Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.
Example:
1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 $\mathrm{g}$ of $\mathrm{NaCI}$ in 1 litre of watcr at $25^{\circ} \mathrm{C}$.
2. An unsaturated solution has the capacity to dissolve more of the compound. $36 \mathrm{~g}$ of $\mathrm{NaCI}$ in 1 litre of water at $25^{\circ} \mathrm{C}$.
3. A super saturated solution is the solution in which crystals can start growing. $500 \mathrm{~g}$ of $\mathrm{NaCI}$ in 1 litre of water at $25^{\circ} \mathrm{C}$.

Question 39.
Explain the effect of pressure on the solubility.
Answer:
1. The change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However the solubility of gases generally increases with increase of pressure.
2. According to Le - chatlier's principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more number of gaseous molecules dissolves in the solvent.
3. If pressure increases, solubility of gas also increases.
Question 40.
A sample of $12 \mathrm{M}$ Concentrated hydrochloric acid has a density $1.2 \mathrm{gL}^{-1}$. Calculate the molality.
Answer:
Given:
Molarity $=12 \mathrm{MHCI}$
Density of the solution $=1.2 \mathrm{~g} \mathrm{~L}^{-1}$
In $12 \mathrm{M} \mathrm{HCl}$ solution, there are 12 moles of $\mathrm{HCl}$ in 1 litre of the solution.
Molality $=\frac{\text { No. of moles of solute }}{\text { Mass of solvent (in kg) }}$
Calculate mass of water (solvent)
Mass of 1 litre $\mathrm{HCI}$ solution = density $\mathrm{x}$ volume
$=1.2 \mathrm{gmL}^{-1} \times 1000 \mathrm{~mL}=1200 \mathrm{~g}$
Mass of $\mathrm{LICI}=$ No. of moles of HCI x molar mass of $\mathrm{HCI}$
$=12 \mathrm{~mol} \mathrm{x} 36.5 \mathrm{~g} \mathrm{~mol}^{-1}=438 \mathrm{~g}$
Mass of waler $=$ mass of $\mathrm{HCI}$ solution - mass of $\mathrm{HCI}$
Mass of waler $=1200-438=762 \mathrm{~g}$
Molalily $=\frac{12}{0.762}=15.75 \mathrm{~m}$

Question 41.
A $0.25 \mathrm{M}$ glucose solution at $370.28 \mathrm{~K}$ has approximately the pressure as blood. What is the osmotic pressure of blood?
Solution.
$
\begin{aligned}
& \mathrm{C}=0.25 \mathrm{M} \\
& \mathrm{T}=37 \mathrm{O} .28 \mathrm{~K} \\
& (\pi)_{\mathrm{gIucose}}=\mathrm{CRT} \\
& (\pi)=0.25 \mathrm{~mol} \mathrm{~L}^{-1} \times 0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{morl}^{-1} \times 370.28 \mathrm{~K} \\
& =7.59 \mathrm{~atm}
\end{aligned}
$
Question 42.
Calculate the molality of a solution containing $7.5 \mathrm{~g}$ of glycine $\left(\mathrm{NH}_2-\mathrm{CH}_2-\mathrm{COOH}\right)$ dissolved in $500 \mathrm{~g}$ of water.
Solution:
$
\text { Molality }=\frac{\text { No. of moles of solute }}{\text { Mass of solvent (in } \mathrm{kg} \text { ) }}
$
No. of moles of glycine $=\frac{\text { Mass of glycine }}{\text { Molar mass of glycine }}=\frac{7.5}{75}=0.1$
$
\text { Molality }=\frac{0.1}{0.5 \mathrm{~kg}}=0.2 \mathrm{~m}
$
Question 43.
Which solution has the lower freezing point? $10 \mathrm{~g}$ of methanol $\left(\mathrm{CH}_3 \mathrm{OH}\right)$ in $100 \mathrm{~g}$ of water (or) $20 \mathrm{~g}$ of ethanol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{HO}\right)$ In $200 \mathrm{~g}$ of water.
Solution:
$
\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m} \text { i.e. } \Delta \mathrm{T}_{\mathrm{f}} \propto \mathrm{m} \\
& \mathrm{m}_{\mathrm{CH}_3 \cdot \mathrm{OH}}=\frac{\left(\frac{10}{32}\right)}{0.1}=3.125 \mathrm{~m} \\
& \mathrm{~m}_{\mathrm{C}_2 \mathrm{H}_5-\mathrm{OH}}=\frac{\left(\frac{20}{46}\right)}{0.2}=2.174 \mathrm{~m} \\
&
\end{aligned}
$
$\therefore$ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Question 44.
How many moles of solute particles are present in one litre of $10^{-4} \mathrm{M}$ potassium sulphate?
Solution:
In $10^{-4} \mathrm{M} \mathrm{K}_2 \mathrm{SO}_4$ solution, there are $10^{-4}$ moles of potassium sulphate.
$\mathrm{K}_2 \mathrm{SO}_4$ molecule contains 3 ions $\left(2 \mathrm{~K}^{+}\right.$and $\left.1 \mathrm{SO}_4{ }^{2-}\right)$
1 mole of $\mathrm{K}_2 \mathrm{SO}_4$ contains $3 \times 6.023 \times 10^{23}$ ions
10 mole of $\mathrm{K}_2 \mathrm{SO}_4$ contains $3 \times 6.023 \times 10^2 \times 10^{-4}$ ions $=18.069 \times 10^{19}$
Question 45.
Henry's law constant for solubility of methane in benzene is $4.2 \times 10^{-5} \mathrm{~mm}$ Hg at a particular constant temperature. At this temperature, calculate the solubiiitv of methane at
1. $750 \mathrm{~mm} \mathrm{Hg}$
2. $84 \mathrm{OnimHg}$
Solution:
$\left(\mathrm{K}_{\mathrm{H}}\right)_{\text {Benzene }}=4.2 \times 10^{-5} \mathrm{~mm} \mathrm{Hg}$. Solubility of methane $=? \mathrm{P}=750 \mathrm{~mm} \mathrm{Hg}, \mathrm{p}=840 \mathrm{~mm} \mathrm{Hg}$
According to Henry's Law,
$\mathrm{P}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{x}_{\text {solution }}$
$750 \mathrm{~mm} \mathrm{Hg}=4.2 \times 10^{-5} \mathrm{~mm} \mathrm{Hg} \cdot \mathrm{x}_{\text {solution }}$
$\mathrm{m}=\frac{\Delta \mathrm{T}_f}{\mathrm{~K}_f}=\frac{0.093 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol} \mathrm{~kg}^{-1}=0.05 \mathrm{~m}$
Question 46.
The observed depression in freezing point of water for a particular solution is $0.093^{\circ} \mathrm{C}$. Calculate the concentration of the solution in molality. Given that molal depression constant for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-}$ 1
Solution:
$
\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\
& \mathrm{m}=\frac{\Delta \mathrm{T}_f}{\mathrm{~K}_f}=\frac{0.093 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol} \mathrm{~kg}^{-1}=0.05 \mathrm{~m} \\
&
\end{aligned}
$
Question 47.
The vapour pressure of pure henzene $\left(\mathrm{C}_6 \mathrm{H}_6\right)$ at a given temperature is $640 \mathrm{~mm} \mathrm{Hg} .2 .2 \mathrm{~g}$ of non-volatile solute is added to $40 \mathrm{~g}$ of benzene. The vapour pressure of the solution is $600 \mathrm{~mm} \mathrm{Hg}$. Calculate the molar mass of the solute?
Solution:
$
\mathrm{P}^0{ }_{\mathrm{C}_6 \mathrm{H}_6}=640 \mathrm{~mm} \mathrm{Hg}
$

$
\begin{aligned}
& \mathrm{W}_2=2.2 \mathrm{~g} \text { (non volatile solute) } \\
& \mathrm{W}_1=40 \mathrm{~g} \text { (benzene) }
\end{aligned}
$
$
\begin{aligned}
& \mathrm{P}_{\text {solution }}=600 \mathrm{~mm} \mathrm{Hg} \\
& \mathrm{M}_2=? \\
& \frac{\mathrm{P}^0-\mathrm{P}}{\mathrm{P}^0}=x_2 \\
& \frac{640-600}{640}=\frac{n_2}{n_1+n_2}\left[\because n_1 \gg n_2 ; n_1+n_2 \approx n_1\right] \\
& \frac{40}{640}=\frac{n_2}{n_1} \\
& 0.0625=\frac{\mathrm{W}_2 \times \mathrm{M}_1}{\mathrm{M}_2 \times \mathrm{W}_1} \\
& \mathrm{M}_2=\frac{2.2 \times 78}{0.0625 \times 40}=68.64 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Evaluate Yourself
Question 1.
If $5.6 \mathrm{~g}$ of $\mathrm{KOH}$ is present in (a) $500 \mathrm{~mL}$ and (b) I litre of solution, calculate the molarity of each of these solutions.
Solution.
Mass of $\mathrm{KOH}=5.6 \mathrm{~g}$
No. of moles $=\frac{5.6}{5.6}=0.1 \mathrm{~mol}$
1. Volume of the solution $=500 \mathrm{ml}=0.5 \mathrm{~L}$
2. Volume of the solution $=$ IL
Molarity $=\frac{\text { Number of moles of solute }}{\text { Volume of solution (in L) }}=\frac{0.1}{0.5}=0.2 \mathrm{M}$

3. Volume of the solution $=\mathrm{IL}$
Molarity $=\frac{0.1}{1} \mathrm{M}$
Question 2.
$2.82 \mathrm{~g}$ of glucose is dissolved in $30 \mathrm{~g}$ of water. Calculate the mole fraction of glucose and water.
Solution:
Mass of glucose $=2.82 \mathrm{~g}$
No. of moles of glucose $=\frac{2.82}{180}=0.016$
Mass of water $=30 \mathrm{~g}=\frac{30}{18}=1.67$
$\mathrm{x}_{\mathrm{H}_2 \mathrm{O}}=\frac{1.67}{1.67+0.016}=\frac{1.67}{1.686}=0.99$
$\mathrm{x}_{\mathrm{H}_2 \mathrm{O}}+\mathrm{x}_{\text {glucose }}=1$
$0.99+\mathrm{x}_{\text {glucose }}=1$
$\mathrm{x}_{\text {glucose }}=1-0.99=0.01$
Question 3.
The antiseptic solution of iodopovidone for the use of external application contains $10 \% \mathrm{w} / \mathrm{v}$ of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of $1.5 \mathrm{~mL}$.
Solution:
$10 \% \frac{w}{v}$ means that $10 \mathrm{~g}$ of solute in $100 \mathrm{ml}$ solution
$\therefore$ Amount of iodopovidone in $1.5 \mathrm{ml}=\frac{10 \mathrm{~g}}{100 \mathrm{ml}} \times 1.5 \mathrm{ml}=0.15 \mathrm{~g}$
Question 4.
A litre of sea water weighing about $1.05 \mathrm{~kg}$ contains $5 \mathrm{mg}$ of dissohed oxygen $\left(\mathrm{O}_2\right)$. Express the concentration of dissolved oxygen in ppm.
Solution:
$
\begin{aligned}
& \frac{\text { Mass of dissolved solid }}{\text { Mass of water }} \times 10^6 \\
& \frac{5 \times 10^{-3} g}{1.05 \times 10^3 g} \times 10^6=4.76 \mathrm{ppm}
\end{aligned}
$

Question 5.
Describe how would you prepare the following solution from pure solute and solvent
1. $1 \mathrm{~L}$ of aqueous solution of $1.5 \mathrm{M} \mathrm{COCI}_2$.
2. $500 \mathrm{~mL}$ of $6.0 \%(\mathrm{v} / \mathrm{v})$ aqueous methanol solution.
Solution:
1. mass of 1.5 moles of $\mathrm{COCI}_2=1.5 \times 129.9=194.85 \mathrm{~g}$
2. $194.85 \mathrm{~g}$ anhydrous cobalt chloride is dissolved in water and the solution is make up to one litre in standard flask.
Question 6.
How much volume of $6 \mathrm{M}$ solution of $\mathrm{NaOH}$ is required to prepare $500 \mathrm{~mL}$ of $0.250 \mathrm{M} \mathrm{NaOH}$ solution. Solution:
$6 \% \frac{v}{v}$ aqueous solution contains $6 \mathrm{~g}$ of methanol in $100 \mathrm{ml}$ solution. To prepare $500 \mathrm{ml}$ of $6 \% \mathrm{v} / \mathrm{v}$ solution of methanol $30 \mathrm{~g}$ methanol is taken in a $500 \mathrm{ml}$ standard flask and required quantity of water is added to make up the solution to $500 \mathrm{ml}$.
Question 7.
Calculate the proportion of $\mathrm{O}_2$ and $\mathrm{N}_2$ dissolved in water at $298 \mathrm{~K}$. When air containing $20 \% \mathrm{O}_2$ and $80 \%$ $\mathrm{N}_2$ by volume is in equilibrium with water at $1 \mathrm{~atm}$ pressure. Henry's law constants for two gases are $\mathrm{KH}\left(\mathrm{O}_2\right)=4.6 \mathrm{x}$ atm and $\mathrm{K}_{\mathrm{H}}\left(\mathrm{N}_2\right) 8.5 \times 10^4 \mathrm{~atm}$.
Solution:
$
\begin{aligned}
& \mathrm{C}_1 \mathrm{~V}_1=\mathrm{C}_2 \mathrm{~V}_2 \\
& 6 \mathrm{M}\left(\mathrm{V}_1\right)=0.25 \mathrm{M} \times 500 \mathrm{ml} \\
& \mathrm{V}_1=\frac{0.25 x 500}{6} \\
& \mathrm{~V}_1=20.3 \mathrm{~mL}
\end{aligned}
$

Question 8.
Explain why the aquatic species are more comfortable in cold water during winter season rather than warm water during the summer.
Solution:
Total pressure $=1 \mathrm{~atm}$
$\mathrm{P}_{\mathrm{N}_2}=\left(\frac{80}{100}\right) \times$ Total pressure $=\frac{80}{100} \times 1 \mathrm{~atm}=0.8 \mathrm{~atm}$
$\mathrm{P}_{\mathrm{O}_2}=\left(\frac{20}{100}\right) \times 1=0.2 \mathrm{~atm}$
According to Henry's Law
$\mathrm{P}_{\text {solute }}=\mathrm{K}_{\mathrm{H}} \mathrm{x}_{\text {solute in solution }}$
$\mathrm{P}_{\mathrm{N}_2}=\left(\mathrm{K}_{\mathrm{H}}\right)_{\text {Nitrogen }} \mathrm{x}$ Mole fraction of Nitrogen in solution
$
\begin{aligned}
& \frac{0.8}{8.5 \times 10^4}=x_{\mathrm{N}_2} \\
& x_{\mathrm{N}_2}=9.4 \times 10^{-6} \\
& x_{\mathrm{O}_2}=\frac{0.2}{4.6 \times 10^4}=4.3 \times 10^{-6}
\end{aligned}
$
Question 9.
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing $128 \mathrm{~g}$ of naphthalene with $39 \mathrm{~g}$ of benzene. It is given that the vapour pressure of pure benzene is $50.71 \mathrm{~mm} \mathrm{Hg}$ and the vapour pressure of pure naphthalene is $32.06 \mathrm{~mm} \mathrm{Hg}$ at $300 \mathrm{~K}$.
Solution:
$\mathrm{P}_{\text {pure benzene }}^0=50.71 \mathrm{~mm} \mathrm{Hg}$
$\mathrm{P}^0{ }_{\text {nepthalene }}=32.06 \mathrm{~mm} \mathrm{Hg}$
Number of moles of benzene $=\frac{39}{78}=0.5 \mathrm{~mol}$
Number of moles of naphthalcne $=\frac{128}{128}=1 \mathrm{~mol}$
Mole fraction of benzene $=\frac{0.5}{1.5}=0.33$
Mole fraction of naphthalene $=1-0.33=0.67$
Partial vapour pressure of benzene $=\mathrm{P}^0$ benzene $\mathrm{x}$ Mole fraction of benzene $=50.71 \times 0.33=16.73 \mathrm{~mm} \mathrm{Hg}$
Partial vapour pressure of naphthalene $=32.06 \times 0.67=21.48 \mathrm{~mm} \mathrm{Hg}$
Mole fraction of benzene in vapour phase $=\frac{16.73}{16.73+21.48}=\frac{16.73}{38.21}=0.44$
Mole fraction of naphthalene in vapour phase $=1-0.44=0.56$
Question 10 .
Vapour pressure of a pure liquid $\mathrm{A}$ is 10.0 torr at $27^{\circ} \mathrm{C}$. The vapour pressure is lowered to 9.0 torr on dissolving one grani of $\mathrm{B}$ in $20 \mathrm{~g}$ of $\mathrm{A}$. If the molar mass of $\mathrm{A}$ is $200 \mathrm{~g} \mathrm{~mol}^{-1}$ then calculate the molar mass of $B$.
Solution:
$
\begin{aligned}
& \mathrm{P}_{\mathrm{A}}^0=10 \text { torr } \\
& \mathrm{P}_{\text {solution }}=9 \text { torr }
\end{aligned}
$

$
\begin{aligned}
& \mathrm{W}_{\mathrm{A}}=20 \mathrm{~g} \\
& \mathrm{~W}_{\mathrm{B}}=1 \mathrm{~g} \\
& \mathrm{M}_{\mathrm{A}}=200 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
$
\begin{aligned}
& \mathrm{M}_{\mathrm{B}}=? \\
& \frac{\Delta \mathrm{P}}{\mathrm{P}_{\mathrm{A}}^0}=\frac{\mathrm{W}_{\mathrm{B}} \times \mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}} \\
& \frac{10-9}{10}=\frac{1 \times 2 U 0}{\mathrm{M}_{\mathrm{B}} \times 20} \\
& \mathrm{M}_{\mathrm{B}}=\frac{200}{20} \times 10=100 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 11.
$2.56 \mathrm{~g}$ of Sulphur is dissolved in $100 \mathrm{~g}$ of carbon disuiphide. The solution boils at $319.692 \mathrm{~K}$. What is the molecular formula ofSulphur in solution? The boiling pointof $\mathrm{CS}_2$ is $319.450 \mathrm{~K}$. Given that $\mathrm{Kb}$ for $\mathrm{CS}_2=$ $2.42 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Solution:
$
\begin{aligned}
& \mathrm{W}_2=2.56 \mathrm{~g} \\
& \mathrm{~W}_1=100 \mathrm{~g} \\
& \mathrm{~T}=319.692 \mathrm{~K} \\
& \mathrm{~K}_{\mathrm{b}}=2.42 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{T}_{\mathrm{b}}=(319.692-319.450) \mathrm{K}=0.242 \mathrm{~K} \\
& \mathrm{M}_2=\frac{\mathrm{K}_b \times \mathrm{W}_2 \times 1000}{\Delta \mathrm{T}_b \times \mathrm{W}_1}=\frac{2.42 \times 2.56 \times 1000}{0.242 \times 100} \\
& \mathrm{M}_2=256 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Molecular mass of sulphur in solulion $=256 \mathrm{~g} \mathrm{~mol}^{-1}$
Atomic mass of one mole of sulphur atom $=32$
No. of atoms in a molecule of sulphur $=\frac{256}{32}=8$
Hence, molecular tòrmula of sulphur is $\mathrm{S}_8$.

Question 12 .
$2 \mathrm{~g}$ of a non electrolyte solute dissolved in $75 \mathrm{~g}$ of benzene lowered the freezing point of benzene by 0.20 $\mathrm{K}$. The freezing point depression constant of benzene is $5.12 \mathrm{~K} \mathrm{Kg} \mathrm{mol}^{-1}$. Find the molar mass of the solute.
Solution:
$
\begin{aligned}
& \mathrm{W}_2=2 \mathrm{~g} \\
& \mathrm{~W}_1=75 \mathrm{~g} \\
& \Delta \mathrm{T}_{\mathrm{f}}=0.2 \mathrm{~K} \\
& \mathrm{k}_{\mathrm{f}}=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}-1 \\
& \mathrm{M}_2=? \\
& \mathrm{M}_2=\frac{\mathrm{K}_f \times \mathrm{W}_2 \times 1000}{\Delta \mathrm{T}_f \times \mathrm{W}_1}=\frac{5.12 \times 2 \times 1000}{0.2 \times 75}=682.66 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 13.
What is the mass of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ in it one litre solution is isotonic with $6 \mathrm{~g} \mathrm{~L}^{-1}$ of urea $\left(\mathrm{NH}_2 \mathrm{CONH}_2\right)$ ?
Solution:
Osmotic pressure of urea solution $\left(\pi_1\right)=C R T$
$
\frac{W_2}{M_2 V} \mathrm{RT}=\frac{6}{60 x 1} \times \mathrm{RT}
$
Osmotic pressure of glucose solution
$
\left(\pi_2\right) \frac{W_2}{180 \times 1} \times \mathrm{RT}
$
For isotonic solution, $\pi_1=\pi_2$
$
\begin{aligned}
& \frac{6}{60}=\frac{W_2}{180 \times 1} \mathrm{RT} \Rightarrow \mathrm{W}_2=\frac{6}{60} \times 180 \\
& \mathrm{~W}_2=18 \mathrm{~g}
\end{aligned}
$
Question 14.
$0.2 \mathrm{~m}$ aqueous solution of $\mathrm{KCI}$ freezes at $-0.68^{\circ} \mathrm{C}$ calculate van't Hoff factor. $\mathrm{K}_{\mathrm{f}}$ for water is $1.86 \mathrm{~K} \mathrm{~kg}$ $\mathrm{mol}^{-1}$.

Solution:
$
i=\frac{\text { Observed property }}{\text { Theoretical property (calculated) }}
$
Given,
$
\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{f}}=0.680 \mathrm{~K} \\
& \mathrm{~m}=0.2 \mathrm{~m}, \\
& \left.\Delta \mathrm{T}_{\mathrm{f}} \text { (observed }\right)=0.680 \mathrm{~K} \\
& \Delta \mathrm{T}_{\mathrm{f}}(\text { Calculated })=\mathrm{k}_{\mathrm{f}} \\
& \mathrm{m}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.2 \mathrm{~mol} \mathrm{~kg}^{-1}=0.372 \mathrm{~K} \\
& i=\frac{\left(\Delta \mathrm{T}_f\right) \text { observed }}{\left(\Delta \mathrm{T}_f\right) \text { calculated }}=\frac{0.680 \mathrm{~K}}{0.372 \mathrm{~K}}=1.82
\end{aligned}
$
Example problems Solved
Question 1.
What volume of $4 \mathrm{M} \mathrm{HCI}$ and $2 \mathrm{M} \mathrm{HCI}$ should be mixed to get $500 \mathrm{~mL}$ of $2 . \mathrm{SM} \mathrm{HCI}$ ?
Solution:
Let the volume of $4 \mathrm{M} \mathrm{HCl}$ required to prepare $500 \mathrm{~mL}$ of $2.5 \mathrm{M} \mathrm{HCI}=\mathrm{x} \mathrm{mL}$
Therefore, the required volume of $2 \mathrm{M} \mathrm{HCI}=(500-\mathrm{x}) \mathrm{mL}$
We know from the equation $\mathrm{x}=\frac{250}{2}=125 \mathrm{~mL}$
Hence, volume of $4 \mathrm{M} \mathrm{HCI}$ required $=125 \mathrm{~mL}$
Volume of $2 \mathrm{M} \mathrm{HCl}$ required $=(500-125) \mathrm{mL}=375 \mathrm{~mL}$
Question 2.
$0.24 \mathrm{~g}$ of a gas dissolves in $1 \mathrm{~L}$ of water at $1.5 \mathrm{~atm}$ pressure. Calculate the amount of dissolved gas when the pressure is raised to $6.0 \mathrm{~atm}$ at constant temperature.
Solution:
$\mathrm{P}_{\text {solute }}=\mathrm{K}_{\mathrm{H}} \mathrm{x}_{\text {solute in solution }}$
At pressure $1.5 \mathrm{~atm}, \mathrm{p}_1=\mathrm{K}_{\mathrm{H}} \mathrm{x}_1$

At pressure $6.0 \mathrm{~atm}, \mathrm{p}_2=\mathrm{K}_{\mathrm{H}^{\mathrm{x}}}$
Dividing equation (1) by (2)
Weget
$
\begin{aligned}
& \frac{P_1}{P_2}=\frac{x_1}{x_2} \\
& \frac{1.5}{6.0}=\frac{0.24}{x_2}
\end{aligned}
$
Therefore
$
\mathrm{x}_2=\frac{0.24 \times 6.0}{1.5}=0.96 \mathrm{~g} / \mathrm{L}
$
Question 3.
An aqueous solution of $2 \%$ nonvolatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when $\mathrm{P}_{\mathrm{A}}{ }^{\circ}$ is 1.013 bar?
Solution:
$
\frac{\Delta \mathrm{P}}{\mathrm{P}_{\mathrm{A}}^0}=\frac{\mathrm{W}_{\mathrm{B}} \times \mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}
$
In a $2 \%$ solution weight of the solute is $2 \mathrm{~g}$ and solvent is $98 \mathrm{~g}$
$
\begin{aligned}
& \Delta \mathrm{P}=\mathrm{P}_{\mathrm{A}}^0-\mathrm{P}_{\text {solution }}=1.013-1.004 \text { bar }=0.009 \text { bar } \\
& \mathrm{M}_{\mathrm{B}}=\frac{\mathrm{P}_{\mathrm{A}}^0 \times \mathrm{W}_{\mathrm{B}} \times \mathrm{M}_{\mathrm{A}}}{\Delta \mathrm{P} \times \mathrm{W}_{\mathrm{A}}} \\
& \mathrm{M}_{\mathrm{B}}=\frac{2 \times 18 \times 1.013}{(98 \times 0.009)}=41.3 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 4.
$0.75 \mathrm{~g}$ of an unknown substance is dissolved in $200 \mathrm{~g}$ solvent. If the elevation of boiling point is $0.15 \mathrm{~K}$ and molal elevation constant is $7.5 \mathrm{~K} \mathrm{~kg}$ more then, calculate the molar mass of unknown substance.
Solution:
$
\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}=\mathrm{K}_{\mathrm{b}} \times \mathrm{W}_2 \times 1000 / \mathrm{M}_2 \times \mathrm{W}_1 \\
& \mathrm{M}_2=\mathrm{K}_{\mathrm{b}} \times \mathrm{W}_2 \times 1000 / \Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_1 \\
& =7.5 \times 0.75 \times 1000 / 0.15 \times 200=187.5 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 5.
Ethylene glycol $\left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right)$ can be used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. $\mathrm{K}_{\mathrm{f}}$ for water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ and molar mass of ethylene glycol is $62 \mathrm{~g} \mathrm{~mol}^{-1}$.
Solution:
Weight of solute $\left(\mathrm{W}_2\right)=20$ mass percent of solution means $20 \mathrm{~g}$ of ethylene glycol
Weight of solvent (water) $\mathrm{W}_1=100-20=80 \mathrm{~g}$
$
\Delta \mathrm{T}_f=\mathrm{K}_f \mathrm{~m}=\frac{\mathrm{K}_f \times \mathrm{W}_2 \times 1000}{\mathrm{M}_2 \times \mathrm{W}_1}=\frac{1.86 \times 20 \times 1000}{62 \times 80}=7.5 \mathrm{~K}
$

The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e. $7.5 \mathrm{~K}$ lower than the normal freezing point of water $(273-7.5) \mathrm{K}=265.5 \mathrm{~K}$
Question 6.
At $400 \mathrm{~K} 1.5 \mathrm{~g}$ of an unknown substance is dissolved in solvent and the solution is made to $1.5 \mathrm{~L}$. Its osmotic pressure is found to be 0.3 bar. Calculate the molar mass of the unknown substance.
Solution:
$
\begin{aligned}
\text { Molar Mass } & =\frac{\text { Mass of unknown solute } \times \mathrm{RT}}{\text { Osmotic pressure } \times \text { Volume of solution }} \\
& =\frac{1.5 \times 8.314 \times 10^{-2} \times 400}{0.3 \times 1.5}=110.85 \mathrm{gram} \mathrm{mol}^{-1}
\end{aligned}
$
Question 7.
The depression in freezing point is $0.24 \mathrm{~K}$ obtained by dissolving $1 \mathrm{~g} \mathrm{NaCI}$ in $200 \mathrm{~g}$ water. Calculate van't Hoff factor. The molal depression constant is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$.
Solution:
Sol. Molar mass of solute
$
\begin{aligned}
\text { Molar mass of solute } & =\frac{1000 \times \mathrm{K}_f \times \text { Mass of } \mathrm{NaCl}}{\Delta \mathrm{T}_f \times \text { Mass of solvent }} \\
& =\frac{1000 \times 1.86 \times 1}{0.24 \times 200}=38.75 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Theoretical molar mass of $\mathrm{NaCl}$ is $=38.75 \mathrm{~g} \mathrm{~mol}^{-1}$
$
i=\frac{\text { Theoretical molar mass }}{\text { Experimental molar mass }}=\frac{58.5}{38.75}=1.50
$