JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

Liquid $M$ and liquid $N$ form an ideal solution. The vapour pressures of pure liquids $M$ and $N$ are $450$ and $700 mmHg,$ respectively, at the same temperature. Then correct statements is:

( ${x}_{M}=Mole fraction of {'M}^{'} in solution;$

${x}_{N}=Mole fraction of {'N}^{'}in solution;$

${y}_{M}=Mole fraction of {'M}^{'} in vapour phase;$

${y}_{N}=Mole fraction of {'N}^{'} in vapour phase;$ )

Official previous-year question

Held on 9 Apr 2019 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

${P}_{M}^{o}$ is given as $=450 mm of Hg.$

${P}_{N}^{o}$ is given as $=700 mm of Hg.$

We know that ${P}_{M}={x}_{M}\times {P}_{ M}^{O}\rightarrow$ Raoult’s law.

${P}_{N}={x}_{N}\times {P}_{ N}^{O}\rightarrow$ Raoult’s law.

From Dalton’s law ${P}_{M}={y}_{M}\times {P}_{T}$

${P}_{N}={y}_{M}\times {P}_{T}$

So,

${x}_{M}\times {P}_{ M}^{o}={y}_{M}\times {P}_{T} \ldots (i)$

${x}_{M}\times {P}_{N}^{o}={y}_{N}\times {P}_{T} \ldots (ii)$

$\frac{(i)}{(\mathrm{ii})} \frac{{x}_{M}}{{x}_{N}}\times \frac{{P}_{ M}^{o}}{{P}_{ N}^{o}}$ = $\frac{{y}_{M}}{{y}_{N}}$

$\frac{{x}_{M}}{{x}_{N}}=\frac{{y}_{M}}{{y}_{N}}\times \frac{750}{450}$

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

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Options

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

$\frac{{x}_{M}}{{x}_{N}}=\frac{{y}_{M}}{{y}_{N}}$

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

$({x}_{M}-{y}_{M})<({x}_{N}-{y}_{N})$

Solution

${P}_{M}^{o}$ is given as $=450 mm of Hg.$

${P}_{N}^{o}$ is given as $=700 mm of Hg.$

We know that ${P}_{M}={x}_{M}\times {P}_{ M}^{O}\rightarrow$ Raoult’s law.

${P}_{N}={x}_{N}\times {P}_{ N}^{O}\rightarrow$ Raoult’s law.

From Dalton’s law ${P}_{M}={y}_{M}\times {P}_{T}$

${P}_{N}={y}_{M}\times {P}_{T}$

So,

${x}_{M}\times {P}_{ M}^{o}={y}_{M}\times {P}_{T} \ldots (i)$

${x}_{M}\times {P}_{N}^{o}={y}_{N}\times {P}_{T} \ldots (ii)$

$\frac{(i)}{(\mathrm{ii})} \frac{{x}_{M}}{{x}_{N}}\times \frac{{P}_{ M}^{o}}{{P}_{ N}^{o}}$ = $\frac{{y}_{M}}{{y}_{N}}$

$\frac{{x}_{M}}{{x}_{N}}=\frac{{y}_{M}}{{y}_{N}}\times \frac{750}{450}$

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

JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

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JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry