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

The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called

A. Dalton's law

B. Avogadro's law

C. Raoult's law

D. Henry's law

Correct Option is (C)

Currently no explanation available

2. ⇒ (MHT CET 2023 11th May Morning Shift )

What is vapour pressure of a solution containing $1 mol$ of a nonvolatile solute in $36 g$ of water $(P_{1}^{0} = 32 mm Hg)$ ?

A. $8.14 mm Hg$

B. $12.31 mm Hg$

C. $16.08 mm Hg$

D. $21.44 mm Hg$

Correct Option is (D)

$\begin{aligned} n_{2} = 1 mol \\ n_{1} = \frac{36}{18} = 2 mol \end{aligned}$

Relative lowering of vapour pressure

$= \frac{P_{1}^{0} - P_{1}}{P_{1}^{0}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}}$

$\begin{array}{ll} ∴ & \frac{32 mm Hg - P_{1}}{32 mm Hg} = \frac{1}{3} \\ 96 mm Hg - 3 P_{1} = 32 mm Hg \\ 64 mm Hg = 3 P_{1} \\ ∴ & P_{1} = 21.33 mm Hg \approx 21.44 mm Hg \end{array}$

3. ⇒ (MHT CET 2021 21th September Evening Shift )

What is vapour pressure of a solution when $2 mol$ of a non-volatile solute are dissolved in $20 mol$ of water? $(P_{1}^{0} = 32 mm Hg)$

A. $29.1 mm Hg$

B. $12 mm Hg$

C. $6 mm Hg$

D. $9 mm Hg$

Correct Option is (A)

$\begin{aligned} P_{1}^{0} = 32 mm Hg, n_{1} = 20 mol, n_{2} = 2 mol \\ \begin{aligned} ∴ x_{1} & = \frac{n_{1}}{n_{1} + n_{2}} = \frac{20}{20 + 2} = 0.909 \\ P_{1} & = P_{1}^{0} x_{1} = 32 mm Hg \times 0.909 \\ = 29.1 mm Hg \end{aligned} \end{aligned}$

4. ⇒ (MHT CET 2021 21th September Morning Shift )

What is vapour pressure of a solution containing 0.1 mol of non-volatile solute dissolved in 16.2 g of water ? (P $_{1}^{0} = 32$ mm Hg)

A. 21.6 mm Hg

B. 28.8 mm Hg

C. 15.7 mm Hg

D. 18.1 mm Hg

Correct Option is (B)

$\begin{aligned} n_{2} = 0.1 mol, n_{1} = \frac{16.2}{18} = 0.9 mol \\ x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = \frac{0.9}{0.1 + 0.9} = 0.9 \\ \begin{array}{r} P_{1} = P_{1}^{0} x_{1} = 32 mm Hg \times 0.9 \\ = 28.8 mm Hg \end{array} \end{aligned}$

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

If vapour pressure of pure solvent and solution are 240 and 216 mm Hg respectively then mole fraction of solvent in solution is

A. 0.9

B. 0.1

C. 0.6

D. 0.4

Correct Option is (A)

$\begin{aligned} P_{1} = 216 mm Hg, P_{1}^{0} = 240 mm Hg \\ Δ P = P_{1}^{0} - P_{1} = 240 - 216 = 24 mm Hg \end{aligned}$

Now, $\frac{Δ P}{P_{1}^{0}} = x_{2}$

$∴ x_{2} = \frac{24 mm Hg}{240 mm Hg} = 0.1$

Hence, mole fraction of solvent, $x_{1} = 1 - x_{2} = 1 - 0.1 = 0.9$

⇒Solution