Q16 / 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: The mole fraction of B in vapour phase which is in equilibrium with this solution 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: The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.
Correct answer: 0.16
Solution
Molality of B in A is $5$ m, which means $5$ moles of B are dissolved in $1$ kg ($1000$ g) of solvent A.
Number of moles of A, $n_A = \dfrac{1000}{50} = 20$ mol
Mole fraction of A in the liquid phase, $x_A = \dfrac{n_A}{n_A + n_B} = \dfrac{20}{20 + 5} = 0.8$
Mole fraction of B in the liquid phase, $x_B = 1 - 0.8 = 0.2$
According to Raoult's law, the partial vapour pressure of A is:
$P_A = P_A^0 x_A = 105 \times 0.8 = 84$ mm Hg
Given total vapour pressure, $P_T = 100$ mm Hg
Partial vapour pressure of B, $P_B = P_T - P_A = 100 - 84 = 16$ mm Hg
Mole fraction of B in the vapour phase, $y_B = \dfrac{P_B}{P_T} = \dfrac{16}{100} = 0.16$
Answer: $0.16$
Number of moles of A, $n_A = \dfrac{1000}{50} = 20$ mol
Mole fraction of A in the liquid phase, $x_A = \dfrac{n_A}{n_A + n_B} = \dfrac{20}{20 + 5} = 0.8$
Mole fraction of B in the liquid phase, $x_B = 1 - 0.8 = 0.2$
According to Raoult's law, the partial vapour pressure of A is:
$P_A = P_A^0 x_A = 105 \times 0.8 = 84$ mm Hg
Given total vapour pressure, $P_T = 100$ mm Hg
Partial vapour pressure of B, $P_B = P_T - P_A = 100 - 84 = 16$ mm Hg
Mole fraction of B in the vapour phase, $y_B = \dfrac{P_B}{P_T} = \dfrac{16}{100} = 0.16$
Answer: $0.16$
Hence, the correct answer is 0.16.