1. ⇒ (MHT CET 2023 12th May Evening Shift )

If $0.15 m$ aqueous solution of KCI freezes at $- {0.511}^{\circ} C$ , calculate van't Hoff factor of KCI (cryoscopic constant of water is $1.86 K kg {mol}^{- 1})$

A. 1.45

B. 1.26

C. 1.82

D. 3.00

Correct Option is (C)

$\begin{array}{ll} Δ T_{f} = T_{f}^{0} - T_{f} = 0 - (- {0.51}^{\circ} C) = {0.51}^{\circ} C = 0.51 K \\ ∴ & Δ T_{f} = {iK}_{f} m \\ ∴ & i = \frac{Δ T}{K_{f} m} = \frac{0.51 K}{1.86 K kg {mol}^{- 1} \times 0.15 mol {kg}^{- 1}} = 1.82 \end{array}$

2. ⇒ (MHT CET 2023 10th May Morning Shift )

Find the depression in freezing point of solution when 3.2 gram non volatile solute with molar mass $128 gram {mol}^{- 1}$ is dissolved in $80 gram$ solvent if cryoscopic constant of solvent is $4.8 K kg {mol}^{- 1}$ .

A. $3.0 K$

B. $1.5 K$

C. $2.0 K$

D. $2.5 K$

Correct Option is (B)

$\begin{aligned} Δ T_{f} & = \frac{1000 K_{f} W_{2}}{M_{2} W_{1}} \\ = \frac{1000 g {kg}^{- 1} \times 4.8 K kg {mol}^{- 1} \times 3.2 g}{128 g {mol}^{- 1} \times 80 g} \\ = 1.5 K \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Morning Shift )

Which of the following solutions exhibits lowest value of boiling point elevation assuming complete dissociation?

A. $0.1 m {AlCl}_{3}$

B. $0.01 m {MgCl}_{2}$

C. $1 m KCl$

D. $0.5 m NaCl$

Correct Option is (B)

	Solution	Moles of particles in 1 kg solution
(A)	0.1 m ${AlCl}_{3}$	0.4
(B)	0.01 m ${MgCl}_{2}$	0.03
(C)	1 m $KCl$	2
(D)	0.5 m $NaCl$	1

$0.01 m {MgCl}_{2}$ solution has minimum number of particles in solution, so it shows the lowest value of boiling point elevation.

4. ⇒ (MHT CET 2023 9th May Evening Shift )

A solution of nonvolatile solute is obtained by dissolving $1 g$ in $100 g$ solvent, decreases its freezing point by $0.3 K$ . Calculate cryoscopic constant of solvent if molar mass of solute is $60 g {mol}^{- 1}$ .

A. $1.0 K kg {mol}^{- 1}$

B. $1.4 K kg {mol}^{- 1}$

C. $2.4 K kg {mol}^{- 1}$

D. $1.8 K kg {mol}^{- 1}$

Correct Option is (D)

$\begin{aligned} Δ T_{f} & = \frac{1000 K_{f} W_{2}}{M_{2} W_{1}} \\ ∴ K_{f} & = \frac{Δ T_{f} M_{2} W_{1}}{1000 W_{2}} \\ = \frac{0.3 K \times 60 g {mol}^{- 1} \times 100 g}{1000 \times 1 g} \\ = 1.8 K kg {mol}^{- 1} \end{aligned}$

5. ⇒ (MHT CET 2021 20th September Evening Shift )

What is cryoscopic constant of water if $5 g$ of glucose in $100 g$ of water has depression in freezing point $2.15 K$ ? (Molar mass of glucose $= 180$ )

A. $7.74 K kg mol^{- 1}$

B. $0.52 K kg {mol}^{- 1}$

C. $1.32 K kg {mol}^{- 1}$

D. $3.86 K kg {mol}^{- 1}$

Correct Option is (A)

$\begin{aligned} W_{2} = 5 g, W_{1} = 100 g \\ M_{2} = 180 g, Δ T_{f} = 2.15 K \\ K_{f} = ? \\ \begin{aligned} Δ T_{f} & = K_{f} \frac{1000 W_{2}}{M_{2} W_{1}} \\ ∴ K_{f} & = \frac{Δ T_{f} M_{2} W_{1}}{1000 W_{2}} = \frac{2.15 \times 180 \times 100}{1000 \times 5} \\ = 7.74 K kg {mol}^{- 1} \end{aligned} \end{aligned}$

⇒Solution