Liquid ' M ' and liquid ' N ' form an ideal solution. The vapour pressures of pure liquids ' M ' and ' N ' are 450 and 700 mm Hg , respectively at the same temperature. Then correct statement is: ( $\mathrm{X}_{\mathrm{M}}=$ mole fraction of ' M ' in solution; $\mathrm{X}_{\mathrm{N}}=$ mole fraction of ' N ' in solution; $\mathrm{y}_{\mathrm{M}}=$ mole fraction of ' M ' in vapour phase; $\mathrm{y}_{\mathrm{N}}=$ mole fraction of ' M ' in vapour phase)

(A) $\frac{x_{M}}{x_{N}}>\frac{y_{M}}{y_{N}}$

(B) $\frac{x_{M}}{x_{N}}=\frac{y_{M}}{y_{N}}$

(C) $\left(x_{M}-y_{M}\right)<\left(x_{N}-y_{N}\right)$

(D) $\frac{x_{M}}{x_{N}}<\frac{y_{M}}{y_{N}}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Solutions

Liquid ' M ' and liquid ' N ' form an ideal solution. The vapour pressures of pure liquids ' M ' and ' N ' are 450 and 700 mm Hg , respectively at the same temperature. Then correct statement is:
(

X_{M} =

mole fraction of ' M ' in solution;

X_{N} =

mole fraction of ' N ' in solution;

y_{M} =

mole fraction of ' M ' in vapour phase;

y_{N} =

mole fraction of ' M ' in vapour phase)

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

Answer: (A)

Solution

By Raoult's law the partial pressures are

P_{M} = x_{M} \cdot 450

and

P_{N} = x_{N} \cdot 700

, and in the vapour each mole fraction is proportional to its partial pressure, so

\frac{y _{M}}{y _{N}} = \frac{P _{M}}{P _{N}} = \frac{x _{M}}{x _{N}} \cdot \frac{450}{700} = 0.64 \frac{x _{M}}{x _{N}}

. The factor

0.64

is less than one, so

\frac{y _{M}}{y _{N}} < \frac{x _{M}}{x _{N}}

, that is

\frac{x _{M}}{x _{N}} > \frac{y _{M}}{y _{N}}

: the vapour is the richer in the more volatile N.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Solutions

Solution

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