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Multiple Choice Questions

1. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^\circ\text{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is: (take $K_b = 0.76\text{ K kg mol}^{-1}$) [JEE 2012 P-II]

(a) 724

(b) 740

(d) 718

Answer: (a)

Explanation:

$$\Delta T_b = K_b \times m = K_b \times \frac{W_s \times 1000}{M_s \times W_{\text{solvent}}}$$

$$2 = \frac{0.76 \times 2.5 \times 1000}{M_s \times 100}$$

$$M_s = \frac{0.76 \times 2.5 \times 1000}{100 \times 2} = 9.5\text{ g mol}^{-1}$$

$$\frac{P^\circ - P_s}{P^\circ} = \frac{n_s}{n_{\text{solv}}} = \frac{W_s / M_s}{W_{\text{solv}} / M_{\text{solv}}} = \frac{2.5 / 9.5}{100 / 18} = \frac{36.0}{100}$$

$$760 - P_s = \frac{36.0}{100} \times 760 \approx 36$$

$$P_s = 760 - 36 = 724\text{ mm Hg}$$

2. The vapour pressure of acetone at $20^\circ\text{C}$ is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at $20^\circ\text{C}$, its vapour pressure was 183 torr. The molar mass ($\text{g mol}^{-1}$) of the substance is: [JEE Main 2015]

(a) 32

(b) 64

(d) 488

Answer: (b)

Explanation:

$$\frac{P^\circ - P_s}{P^\circ} = \frac{n_{\text{solute}}}{n_1 + n_2}$$

$$\frac{185 - 183}{185} = \frac{1.2 / M}{100 / 58}$$

$$\frac{2}{185} = \frac{1.2 \times 58}{M \times 100}$$

$$\Rightarrow M = 64\text{ g mol}^{-1}$$

3. 18 g glucose is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is : [JEE Main 2016]

(a) 7.6

(b) 76.0

(d) 759.0

Answer: (c)

Explanation: Moles of glucose $= \frac{18}{180} = 0.1$

Moles of water $= \frac{178.2}{18} = 9.9$

$\Rightarrow \text{Total } n = 10$

$$\frac{\Delta P}{P^\circ} = \frac{0.1}{10}$$

$$\Delta P = 0.01 P^\circ = 0.01 \times 760 = 7.6\text{ torr}$$

$$P_s = 760 - 7.6 = 752.4\text{ torr}$$

4. Mixture(s) showing positive deviation from Raoult’s law at $35^\circ\text{C}$ is(are): [JEE Adv. 2016 P-II]

(a) carbon tetrachloride + methanol

(b) carbon disulphide + acetone

(d) phenol + aniline

Answer: (a, b)

Explanation:

(a) $\text{CCl}_4 + \text{CH}_3\text{OH} \implies$ Positive deviation from Raoult's law

(b) $\text{CS}_2 + \text{CH}_3\text{COCH}_3 \implies$ Positive deviation from Raoult's law

(d) Phenol + Aniline $\implies$ Negative deviation from Raoult's law.

5. The freezing point of benzene decreases by $0.45^\circ\text{C}$ when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: ($K_f \text{ for benzene} = 5.12\text{ K kg mol}^{-1}$) [JEE Main 2017]

(a) 94.6%

(b) 64.6%

(d) 74.6%

Answer: (a)

Explanation: In benzene, $2\text{CH}_3\text{COOH} \longrightarrow (\text{CH}_3\text{COOH})_2$

$$i = 1 + \left(\frac{1}{2} - 1\right)\alpha = 1 - \frac{\alpha}{2}$$

here, $\alpha$ is degree of association.

$$\Delta T_f = i K_f m$$

$$0.45 = \left(1 - \frac{\alpha}{2}\right) (5.12) \frac{(0.2 / 60)}{20 / 1000}$$

$$1 - \frac{\alpha}{2} = 0.527 \implies \frac{\alpha}{2} = 0.473 \implies \alpha = 0.946$$

degree of association = 94.6%

6. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here $x_L$ and $x_M$ represent mole fractions of L and M, respectively, in the solution. The correct statement(s) applicable to this system is(are): [JEE Adv. 2017 P-I]

(a) Attractive intermolecular interactions between L – L in pure liquid L and M – M in pure liquid M are stronger than those between L – M when mixed in solution

(b) The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when $x_L \to 0$

(d) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when $x_L \to 1$

Answer: (a, d)

Explanation: Attractive intermolecular interactions between L – L in pure liquid L and M – M in pure liquid M are stronger than those between L – M when mixed in solution. The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when $x_L \to 1$.

7. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point? [JEE Main 2018]

(a) $[\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O}$

(b) $[\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_3$

(d) $[\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}$

Answer: (a)

Explanation: The solution which shows maximum freezing point must have minimum number of solute particles.

(a) $[\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O} \longrightarrow [\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3]$, $i = 1$

(b) $[\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_3 \longrightarrow [\text{Co}(\text{H}_2\text{O})_6]^{3+} + 3\text{Cl}^-$, $i = 4$

(c) $[\text{Co}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O} \longrightarrow [\text{Co}(\text{H}_2\text{O})_5\text{Cl}]^{2+} + 2\text{Cl}^-$, $i = 3$

(d) $[\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O} \longrightarrow [\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]^+ + \text{Cl}^-$, $i = 2$

So, Solution of 1 molal $[\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O}$ will have minimum number of particles in aqueous state. Hence, option (a) is correct.

8. An aqueous solution contains an unknown concentration of $\text{Ba}^{2+}$. When 50 mL of a 1 M solution of $\text{Na}_2\text{SO}_4$ is added, $\text{BaSO}_4$ just begins to precipitate. The final volume is 500 mL. The solubility product of $\text{BaSO}_4$ is $1 \times 10^{-10}$. What is original concentration of $\text{Ba}^{2+}$? [JEE Main 2018]

(a) $1 \times 10^{-10}\text{ M}$

(b) $5 \times 10^{-9}\text{ M}$

(d) $1.1 \times 10^{-9}\text{ M}$

Answer: (d)

Explanation: Final concentration of $[\text{SO}_4^{2-}] = \frac{50 \times 1}{500} = 0.1\text{ M}$

$K_{sp} \text{ of BaSO}_4 = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = 1 \times 10^{-10}$

$$[\text{Ba}^{2+}] \times 0.1 = 10^{-10} \implies [\text{Ba}^{2+}]_{\text{final}} = 10^{-9}\text{ M}$$

Concentration of $\text{Ba}^{2+}$ in original solution:

$$M_1 V_1 = M_2 V_2$$

$$M_1 (500 - 50) = 10^{-9} (500)$$

$$M_1 = \frac{500}{450} \times 10^{-9} = 1.11 \times 10^{-9}\text{ M}$$

So, option (d) is correct.

9. When metal ‘M’ is treated with $\text{NaOH}$, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of $\text{NaOH}$. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is: [JEE Main 2018]

(a) Fe

(b) Zn

(d) Al

Answer: (d)

Explanation:

$$\text{Al} \xrightarrow{\text{NaOH}} \text{Al(OH)}_3 \downarrow \text{ (White gelatinous ppt.)} \xrightarrow{\text{excess NaOH}} \text{NaAlO}_2 \text{ (Sodium meta aluminate, soluble)}$$

$$2\text{Al(OH)}_3 \xrightarrow{\text{Strong heating}} \text{Al}_2\text{O}_3 + 3\text{H}_2\text{O}$$

$\text{Al}_2\text{O}_3$ is used in column chromatography.

10. An aqueous solution contains 0.10 M $\text{H}_2\text{S}$ and 0.20 M $\text{HCl}$. If the equilibrium constant for the formation of $\text{HS}^-$ is $1.0 \times 10^{-7}$ and that of $\text{S}^{2-}$ from $\text{HS}^-$ ions is $1.2 \times 10^{-13}$, then the concentration of $\text{S}^{2-}$ ions in aqueous solution is: [JEE Main 2018]

(a) $5 \times 10^{-19}$

(b) $5 \times 10^{-8}$

(d) $6 \times 10^{-21}$

Answer: (c)

Explanation: In presence of external $\text{H}^+$:

$$\text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-}, \quad K_{eq} = K_{a1} \times K_{a2}$$

$$K_{eq} = \frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]} = 1.0 \times 10^{-7} \times 1.2 \times 10^{-13}$$

$$\frac{[0.2]^2 [\text{S}^{2-}]}{[0.1]} = 1.2 \times 10^{-20}$$

$$[\text{S}^{2-}] = \frac{1.2 \times 10^{-20} \times 0.1}{0.04} = 3 \times 10^{-20}\text{ M}$$

11. The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively are: [JEE Main 2019]

(a) 500 mmHg, 0.5, 0.5

(b) 450 mmHg, 0.4, 0.6

(d) 500 mmHg, 0.4, 0.6

Answer: (d)

Explanation: $P_{\text{total}} = X_A P_A^\circ + X_B P_B^\circ = 0.5 \times 400 + 0.5 \times 600 = 500\text{ mmHg}$

Now, mole fraction of A in vapour phase:

$$Y_A = \frac{P_A}{P_{\text{total}}} = \frac{0.5 \times 400}{500} = 0.4$$

and mole fraction of B in vapour phase:

$$Y_B = 1 - 0.4 = 0.6$$

12. If solubility product of $\text{Zr}_3(\text{PO}_4)_4$ is denoted by $K_{sp}$ and its molar solubility is denoted by S, then which of the following relation between S and $K_{sp}$ is correct. [JEE Main 2019]

(a) $S = \left(\frac{K_{sp}}{929}\right)^{1/9}$

(b) $S = \left(\frac{K_{sp}}{216}\right)^{1/7}$

(d) $S = \left(\frac{K_{sp}}{6912}\right)^{1/7}$

Answer: (d)

Explanation: $\text{Zr}_3(\text{PO}_4)_4 (s) \rightleftharpoons 3\text{Zr}^{4+} (aq) + 4\text{PO}_4^{3-} (aq)$

$$K_{sp} = [\text{Zr}^{4+}]^3 [\text{PO}_4^{3-}]^4 = (3S)^3 \cdot (4S)^4 = 27 S^3 \cdot 256 S^4 = 6912 S^7$$

$$\therefore S = \left(\frac{K_{sp}}{6912}\right)^{1/7}$$

13. For the solution of the gases w, x, y and z in water at 298K, the Henry’s law constants ($K_H$) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is: [JEE Main 2019]

(a) Graph showing partial pressure vs mole fraction with slopes z > y > x > w

(b) Graph showing partial pressure vs mole fraction

(d) Graph showing partial pressure vs mole fraction

Answer: (c)

Explanation: According to Henry's law: $P = K_H \times \chi_{\text{gas}}$. Higher the value of $K_H$, steeper is the slope of $P$ vs $\chi_{\text{gas}}$ plot. Therefore, slope for z > y > x > w.

14. Which one of the following statements regarding Henry’s law not correct? [JEE Main 2019]

(a) The value of $K_H$ increases with function of the nature of the gas.

(b) Higher the value of $K_H$ at a given pressure, higher is the solubility of the gas in the liquids.

(d) Different gases have different $K_H$ (Henry’s law constant) values at the same temperature.

Answer: (b)

Explanation: $P = K_H \times X_{\text{gas}}$. More is $K_H$, less is solubility. Higher temperature leads to higher $K_H$ and lower solubility. Hence statement (b) is incorrect.

15. A solution of sodium sulfate contains 92 g of $\text{Na}^+$ ions per kilogram of water, the molality of $\text{Na}^+$ ions in that solution in $\text{mol kg}^{-1}$ is: [JEE Main 2019]

(a) 16

(b) 8

(d) 12

Answer: (c)

Explanation: $n_{\text{Na}^+} = \frac{92}{23} = 4\text{ moles}$

$$\text{Molality} = \frac{4\text{ moles}}{1\text{ kg water}} = 4\text{ mol kg}^{-1}$$

16. Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressures of pure A and pure B are $7 \times 10^3\text{ Pa}$ and $12 \times 10^3\text{ Pa}$, respectively. The composition of the vapor in equilibrium with a solution containing 40 mole percent of A at this temperature is: [JEE Main 2019]

(a) $x_A = 0.37; x_B = 0.63$

(b) $x_A = 0.28; x_B = 0.72$

(d) $x_A = 0.4; x_B = 0.6$

Answer: (b)

Explanation:

$$x_A = \frac{P_A^\circ X_A}{P_A^\circ X_A + P_B^\circ X_B} = \frac{7 \times 10^3 \times 0.4}{7 \times 10^3 \times 0.4 + 12 \times 10^3 \times 0.6}$$

$$x_A = \frac{2.8}{2.8 + 7.2} = \frac{2.8}{10} = 0.28 \implies x_B = 0.72$$

17. The amount of sugar ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$) required to prepare 2 L of its 0.1 M aqueous solution is: [JEE Main 2019]

(a) 68.4 g

(b) 17.1 g

(d) 136.8 g

Answer: (a)

Explanation: $\text{Molarity} = \frac{n}{V (\text{in L})} \implies 0.1 = \frac{\text{wt} / 342}{2}$

$$\text{wt}(\text{C}_{12}\text{H}_{22}\text{O}_{11}) = 0.1 \times 2 \times 342 = 68.4\text{ g}$$

18. Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point of 2 molal solutions of glucose in the same solvent is 2 K. The relation between $K_b$ and $K_f$ is: [JEE Main 2019]

(a) $K_b = 0.5 K_f$

(b) $K_b = 2 K_f$

(d) $K_b = K_f$

Answer: (b)

Explanation:

$$\frac{\Delta T_b}{\Delta T_f} = \frac{i \cdot m_1 \cdot K_b}{i \cdot m_2 \cdot K_f}$$

$$\frac{2}{2} = \frac{1 \times 1 \times K_b}{1 \times 2 \times K_f} \implies 1 = \frac{K_b}{2 K_f} \implies K_b = 2 K_f$$

19. The concentration of dissolved oxygen (DO) in cold water can go upto: [JEE Main 2019]

(a) 10 ppm

(b) 14 ppm

(d) 8 ppm

Answer: (a)

Explanation: In cold water, dissolved oxygen (DO) can reach a concentration up to 10 ppm.

20. 8g of $\text{NaOH}$ is dissolved in 18g of $\text{H}_2\text{O}$. Mole fraction of $\text{NaOH}$ in solution and molality (in $\text{mol kg}^{-1}$) of the solutions respectively are: [JEE Main 2019]

(a) 0.167, 11.11

(b) 0.2, 22.20

(d) 0.167, 22.20

Answer: (a)

Explanation: 8 g $\text{NaOH}$, mol of $\text{NaOH} = \frac{8}{40} = 0.2\text{ mol}$

18 g $\text{H}_2\text{O}$, mol of $\text{H}_2\text{O} = \frac{18}{18} = 1\text{ mol}$

$$\therefore X_{\text{NaOH}} = \frac{0.2}{1.2} = 0.167$$

$$\text{Molality} = \frac{0.2 \times 1000}{18} = 11.11\text{ m}$$

21. If $K_{sp}$ of $\text{Ag}_2\text{CO}_3$ is $8 \times 10^{-12}$, the molar solubility of $\text{Ag}_2\text{CO}_3$ in $0.1\text{M AgNO}_3$ is : [JEE Main 2019]

(a) $8 \times 10^{-12}\text{ M}$

(b) $8 \times 10^{-10}\text{ M}$

(d) $8 \times 10^{-13}\text{ M}$

Answer: (b)

Explanation: $\text{Ag}_2\text{CO}_3(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CO}_3^{2-}(aq)$

$$K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}]$$

$$8 \times 10^{-12} = (0.1 + 2S)^2 \cdot S \approx (0.1)^2 \cdot S$$

$$S = \frac{8 \times 10^{-12}}{0.01} = 8 \times 10^{-10}\text{ M}$$

Paragraph

Paragraph for Question Nos. 22 to 23

The electrochemical cell shown below is a concentration cell.

$\text{M}|\text{M}^{2+}\text{ (saturated solution of sparingly soluble salt, }\text{MX}_2)|\text{M}^{2+}(0.001\text{ mol dm}^{-3})|\text{M}$

The emf of the cell depends on the difference in concentration of $\text{M}^{2+}$ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V. [JEE 2012 P-II]

22. The value of $\Delta G$ (in $\text{kJ mol}^{-1}$) for the given cell is (take $1\text{ F} = 96500\text{ C mol}^{-1}$)

(a) $-5.7$

(b) $5.7$

(d) $-11.4$

Answer: (d)

Explanation:

$$\Delta G = -nFE_{\text{cell}} = -2 \times 96500 \times 0.059 = -11387\text{ J mol}^{-1} = -11.4\text{ kJ mol}^{-1}$$

23. The solubility product of $\text{MX}_2$ at 298 K based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059\text{ V}$)

(a) $1 \times 10^{-15}$

(b) $4 \times 10^{-15}$

(d) $4 \times 10^{-12}$

Answer: (b)

Explanation:

$$E = \frac{0.0591}{2} \log \frac{C_2}{C_1} \implies 0.059 = \frac{0.059}{2} \log \frac{10^{-3}}{[\text{M}^{2+}]}$$

$$\log \frac{10^{-3}}{[\text{M}^{2+}]} = 2 \implies \frac{10^{-3}}{[\text{M}^{2+}]} = 100 \implies [\text{M}^{2+}] = 10^{-5}\text{ M}$$

$$K_{sp} = [\text{M}^{2+}][\text{X}^-]^2 = (10^{-5})(2 \times 10^{-5})^2 = 4 \times 10^{-15}\text{ mol}^3\text{ dm}^{-9}$$

Integer and Subjective

24. A compound $\text{H}_2\text{X}$ with molar weight of 80 g dissolved in a solvent having density of $0.4\text{ g ml}^{-1}$. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is: [JEE Adv. 2014 P-I]

Answer: 8

Explanation: Since, in solution, 3.2 moles of solute are present in $1000\text{ mL}$ solution.

Weight of solvent $= 1000 \times 0.4 = 400\text{ g} = 0.4\text{ kg}$

$$\text{Molality} = \frac{3.2}{0.4} = 8\text{ m}$$

25. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is $2.0\text{ g cm}^{-3}$. The ratio of the molecular weights of the solute and solvent, $\left(\frac{\text{MW}_{\text{solute}}}{\text{MW}_{\text{solvent}}}\right)$ is: [JEE Adv. 2016 P-I]

Answer: 9

Explanation: Molality $(m) = \frac{x_{\text{solute}} \times 1000}{x_{\text{solvent}} \times \text{MW}_{\text{solvent}}} = \frac{0.1 \times 1000}{0.9 \times \text{MW}_{\text{solvent}}}$

$\text{Molarity } (M) = \frac{0.1 \times 2 \times 1000}{0.1 \times \text{MW}_{\text{solute}} + 0.9 \times \text{MW}_{\text{solvent}}}$

As Molality = Molarity:

$$\frac{0.1 \times 1000}{0.9 \times \text{MW}_{\text{solvent}}} = \frac{0.1 \times 2 \times 1000}{0.1 \times \text{MW}_{\text{solute}} + 0.9 \times \text{MW}_{\text{solvent}}}$$

$$0.1 \times \text{MW}_{\text{solute}} + 0.9 \times \text{MW}_{\text{solvent}} = 1.8 \times \text{MW}_{\text{solvent}}$$

$$0.1 \times \text{MW}_{\text{solute}} = 0.9 \times \text{MW}_{\text{solvent}} \implies \frac{\text{MW}_{\text{solute}}}{\text{MW}_{\text{solvent}}} = 9$$

26. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions $x_A$ and $x_B$, respectively, has vapour pressure of 22.5 Torr. The value of $x_A/x_B$ in the new solution is ___. (given that the vapour pressure of pure liquid A is 20 Torr at temperature T) [JEE Adv. 2018 P-I]

Answer: 19

Explanation: $P_A^\circ = 20\text{ Torr}$. For equimolar binary solution ($x_A = x_B = 0.5$):

$$\frac{P_A^\circ + P_B^\circ}{2} = 45 \implies 20 + P_B^\circ = 90 \implies P_B^\circ = 70\text{ Torr}$$

For the new solution:

$$P = P_B^\circ + (P_A^\circ - P_B^\circ)x_A \implies 22.5 = 70 + (20 - 70)x_A$$

$$50 x_A = 47.5 \implies x_A = \frac{47.5}{50} = 0.95 \implies x_B = \frac{2.5}{50} = 0.05$$

$$\frac{x_A}{x_B} = \frac{47.5}{2.5} = 19$$

27. The plot given below shows P – T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents. On addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is _____. [JEE Adv. 2018 P-I]

Answer: 0.05

Explanation: When $\text{NaCl}$ as solute is used ($i = 2$):

For solvent X: $\Delta T_b = 2 K_b m$, For solvent Y: $\Delta T_b' = 2 K_b' m$

From the P-T curve: $\frac{K_b}{K_b'} = 2$

When solute S is used ($i = 1 - \frac{\alpha}{2}$):

For solvent Y ($\alpha = 0.7$): $i_Y = 1 - \frac{0.7}{2} = 0.65$

$$\Delta T_b = \left(1 - \frac{\alpha}{2}\right) K_b m, \quad \Delta T_b' = (0.65) K_b' m$$

$$\frac{\Delta T_b}{\Delta T_b'} = \frac{(1 - \alpha/2) K_b}{(0.65) K_b'} = 3$$

$$\left(1 - \frac{\alpha}{2}\right) \times 2 = 3 \times 0.65 \implies 1 - \frac{\alpha}{2} = \frac{1.95}{2} = 0.975$$

$$\frac{\alpha}{2} = 0.025 \implies \alpha = 0.05$$

28. Mole fraction of urea in 900 gram water is 0.05. Density of solution is $1.2\text{ g/cm}^3$. Find molarity of solution. [JEE Adv. 2019 P-II]

Answer: 2.98

Explanation: $\text{wt of H}_2\text{O} = 900\text{ g} \implies n_{\text{H}_2\text{O}} = \frac{900}{18} = 50\text{ moles}$

Let moles of urea $= y$

$$x_{\text{urea}} = \frac{y}{y + 50} = 0.05 \implies y = \frac{50}{19}\text{ moles}$$

$$\text{Wt of urea} = \frac{50}{19} \times 60 = \frac{3000}{19}\text{ g}$$

$$\text{Wt of solution} = 900 + \frac{3000}{19} = \frac{20100}{19}\text{ g}$$

$$\text{Volume of solution} = \frac{\text{wt}}{\rho} = \frac{20100}{19 \times 1.2}\text{ mL}$$

$$\text{Molarity} = \frac{50 / 19}{\frac{20100}{19 \times 1.2}} \times 1000 = \frac{50 \times 1.2 \times 1000}{20100} = \frac{600}{201} \approx 2.985\text{ M} \approx 2.98\text{ M}$$