Q15 / 54 · JEE Advanced 2026 (Paper 2) · Chemistry
Passage: Two volatile liquids A and B form an ideal solution. Consider a $5$ molal solution of B in A inside a closed container having a total vapour pressure of $100$ mm Hg at $300$ K. The vapour pressure of pure A at $300$ K is $105$ mm Hg. Assume that A and B behave as ideal gases in the vapour phase.
Given:
The gas constant $R = 0.08$ L atm K$^{-1}$ mol$^{-1}$
Molar mass of A is $50$ g mol$^{-1}$
Molar mass of B is $57$ g mol$^{-1}$
Density of liquid B at $300$ K is $0.5$ g/mL
$1$ atm $= 760$ mm Hg
Question: At $300$ K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is ____.
Given:
The gas constant $R = 0.08$ L atm K$^{-1}$ mol$^{-1}$
Molar mass of A is $50$ g mol$^{-1}$
Molar mass of B is $57$ g mol$^{-1}$
Density of liquid B at $300$ K is $0.5$ g/mL
$1$ atm $= 760$ mm Hg
Question: At $300$ K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is ____.
Correct answer: 2000
Solution
For a $5$ molal solution of B in A, there are $5$ moles of B in $1000$ g of A.
Moles of A $= \dfrac{1000}{50} = 20$ mol.
Mole fraction of B, $x_B = \dfrac{5}{20 + 5} = 0.2$
Mole fraction of A, $x_A = 1 - 0.2 = 0.8$
Using Raoult's law for the total vapour pressure:
$P_T = P_A^\circ x_A + P_B^\circ x_B$
$100 = 105 \times 0.8 + P_B^\circ \times 0.2$
$100 = 84 + 0.2 P_B^\circ$
$0.2 P_B^\circ = 16 \Rightarrow P_B^\circ = 80$ mm Hg
The molar volume of pure B in the liquid phase ($V_{m,l}$) is:
$V_{m,l} = \dfrac{\text{Molar mass of B}}{\text{Density of liquid B}} = \dfrac{57}{0.5} = 114$ mL/mol $= 0.114$ L/mol
Assuming pure B behaves as an ideal gas in the vapour phase, its molar volume ($V_{m,v}$) at its vapour pressure $P_B^\circ$ is:
$V_{m,v} = \dfrac{RT}{P_B^\circ} = \dfrac{0.08 \times 300}{\left(\dfrac{80}{760}\right)} = \dfrac{24 \times 760}{80} = 228$ L/mol
The ratio of the molar volume of pure B in the vapour phase to its molar volume in the liquid phase is:
$\text{Ratio} = \dfrac{V_{m,v}}{V_{m,l}} = \dfrac{228}{0.114} = 2000$
Answer: $2000$
Moles of A $= \dfrac{1000}{50} = 20$ mol.
Mole fraction of B, $x_B = \dfrac{5}{20 + 5} = 0.2$
Mole fraction of A, $x_A = 1 - 0.2 = 0.8$
Using Raoult's law for the total vapour pressure:
$P_T = P_A^\circ x_A + P_B^\circ x_B$
$100 = 105 \times 0.8 + P_B^\circ \times 0.2$
$100 = 84 + 0.2 P_B^\circ$
$0.2 P_B^\circ = 16 \Rightarrow P_B^\circ = 80$ mm Hg
The molar volume of pure B in the liquid phase ($V_{m,l}$) is:
$V_{m,l} = \dfrac{\text{Molar mass of B}}{\text{Density of liquid B}} = \dfrac{57}{0.5} = 114$ mL/mol $= 0.114$ L/mol
Assuming pure B behaves as an ideal gas in the vapour phase, its molar volume ($V_{m,v}$) at its vapour pressure $P_B^\circ$ is:
$V_{m,v} = \dfrac{RT}{P_B^\circ} = \dfrac{0.08 \times 300}{\left(\dfrac{80}{760}\right)} = \dfrac{24 \times 760}{80} = 228$ L/mol
The ratio of the molar volume of pure B in the vapour phase to its molar volume in the liquid phase is:
$\text{Ratio} = \dfrac{V_{m,v}}{V_{m,l}} = \dfrac{228}{0.114} = 2000$
Answer: $2000$
Hence, the correct answer is 2000.