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

Calculate osmotic pressure of solution of 0.025 mole glucose in $100 mL$ water at $300 K$ . $[R = 0.082 atm dm {mol}^{- 1} K^{- 1}]$

A. $1.54 atm$

B. $2.05 atm$

C. $6.15 atm$

D. $3.08 atm$

Correct Option is (C)

$\begin{aligned} π & = MRT = \frac{n_{2} RT}{V} \\ = \frac{0.025 mol \times 0.082 {dm}^{3} atm {mol}^{- 1} K^{- 1} \times 300 K}{0.1 {dm}^{3}} \\ = 6.15 atm \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

Calculate molality of solution of a nonvolatile solute having boiling point elevation $1.89 K$ if boiling point elevation constant of solvent is $3.15 K kg {mol}^{- 1}$ .

A. 0.4 m

B. 0.8 m

C. 0.6 m

D. 0.3 m

Correct Option is (C)

$\begin{aligned} Δ T_{b} & = K_{b} \times m \\ 1.89 & = 3.15 \times m \\ ∴ m & = \frac{1.89}{3.15} = 0.6 mol {kg}^{- 1} \end{aligned}$

3. ⇒ (MHT CET 2023 11th May Evening Shift )

A solution of nonvolatile solute is obtained by dissolving $1.5 g$ in $30 g$ solvent has boiling point elevation $0.65 K$ . Calculate the molal elevation constant if molar mass of solute is $150 g {mol}^{- 1}$ .

A. $1.95 K kg {mol}^{- 1}$

B. $2.23 K kg {mol}^{- 1}$

C. $1.52 K kg {mol}^{- 1}$

D. $2.72 K kg {mol}^{- 1}$

Correct Option is (A)

$\begin{aligned} M_{2} = \frac{1000 K_{b} W_{2}}{Δ T_{b} W_{1}} \\ K_{b} = \frac{M_{2} \times Δ T_{b} \times W_{1}}{1000 \times W_{2}} & = \frac{150 \times 0.65 \times 30}{1000 \times 1.5} \\ = 1.95 K kg {mol}^{- 1} \end{aligned}$

4. ⇒ (MHT CET 2023 11th May Evening Shift )

Calculate osmotic pressure of $0.2 M$ aqueous $KCl$ solution at $0^{\circ} C$ if van't Hoff factor for $KCl$ is 1.83. $[R = 0.082 {dm}^{3} atm {mol}^{- 1} K^{- 1}]$

A. $8.2 atm$

B. $9.4 atm$

C. $10.6 atm$

D. $6.5 atm$

Correct Option is (A)

$\begin{aligned} π & = iMRT \\ = 1.83 \times 0.2 \times 0.082 \times 273 \\ = 8.2 atm \end{aligned}$

5. ⇒ (MHT CET 2023 10th May Evening Shift )

What is osmotic pressure of solution of $1.7 g {CaCl}_{2}$ in $1.25 {dm}^{3}$ water at $300 K$ if van't Hoff factor and molar mass of ${CaCl}_{2}$ , are 2.47 and $111 g {mol}^{- 1}$ respectively?

$[R = 0.082 {dm}^{3} atm {mol}^{- 1} K^{- 1}]$

A. $0.625 atm$

B. $0.744 atm$

C. $0.827 atm$

D. $0.936 atm$

Correct Option is (B)

$π = iMRT = \frac{i \times W_{2} RT}{M_{2} V}$

$π = \frac{2.47 \times 1.7 g \times 0.082 d m^{3} atm mo l^{- 1} K^{- 1} \times 300 K}{111 g mo l^{- 1} \times 1.25 d m^{3}} = 0.744 atm$

⇒Solution