JEE Main 2025 Apr 7 Shift 2, Chemistry Q9: Vapour Pressure Of A Solution

JEE Main2025Apr 7, Shift 2Chemistry

Liquid A and B form an ideal solution. The vapour pressure of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If $x_{A}$ and $x_{B}$ are the mole fraction of A and B in solution while $y_{A}$ and $y_{B}$ are the mole fraction of A and B in vapour phase then :

A
$\frac{x _{A}}{x _{B}} < \frac{y _{A}}{y _{B}}$
B
$\frac{x _{A}}{x _{B}} = \frac{y _{A}}{y _{B}}$
C
$\frac{x _{A}}{x _{B}} > \frac{y _{A}}{y _{B}}$
D
$(x_{A} - y_{A}) < (x_{B} - y_{B})$

Solution

For an ideal solution, Raoult's law gives partial pressures $p_{A} = x_{A} P_{A}^{\circ}$ and $p_{B} = x_{B} P_{B}^{\circ}$ , and Dalton's law gives the vapour-phase mole fractions as $y_{A} = \frac{p _{A}}{p _{A} + p _{B}}$ and $y_{B} = \frac{p _{B}}{p _{A} + p _{B}}$ .

Dividing, $\frac{y _{A}}{y _{B}} = \frac{x _{A} P _{A}^{\circ}}{x _{B} P _{B}^{\circ}}$ , so $\frac{y _{A} / y _{B}}{x _{A} / x _{B}} = \frac{P _{A}^{\circ}}{P _{B}^{\circ}}$ .

Since $P_{A}^{\circ} = 350$ mm Hg is less than $P_{B}^{\circ} = 750$ mm Hg, this ratio is less than 1, which means $\frac{y _{A}}{y _{B}} < \frac{x _{A}}{x _{B}}$ , i.e. $\frac{x _{A}}{x _{B}} > \frac{y _{A}}{y _{B}}$ .

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JEE Main 2025 Apr 7 Shift 2, Chemistry Q9: Vapour Pressure Of A Solution

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