JEE Main 2017 — Chemistry Physical Chemistry

2017mcqmedium

A solution is prepared by mixing $8.5$ g of ${\mathrm{CH}}_{2}{\mathrm{Cl}}_{2}$ and $11.95 \text{g}$ of ${\mathrm{CHCl}}_{3}$ . If vapour pressure of ${\mathrm{CH}}_{2}{\mathrm{Cl}}_{2}$ and ${\mathrm{CHCl}}_{3}$ at $298$ K are $415$ and $200$ mm Hg respectively, the mole fraction of ${\mathrm{CHCl}}_{3}$ in vapour form is: $(\text{M}\text{o}\text{l}\text{a}\text{r} \text{m}\text{a}\text{s}\text{s} \text{o}\text{f} \mathrm{Cl}=35.5 \text{g} \text{m}\text{o}{\text{l}}^{-1})$

Official previous-year question

Held on 9 Apr 2017 · Verified 6 Jul 2026.

Options

A
$0.162$
B
$0.675$
C
$0.325$
D
$0.486$

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Solution

${\text{mole of CH}}_{2}\text{C}{\text{l}}_{2} \text{in liquid phase}=\frac{8.5}{85}=0.1$

${\text{mole of CHCl}}_{\text{3}} \text{in liquid phase}=\frac{11.95}{119.5}=0.1$

${\text{mole fraction of CH}}_{\text{2}}{\text{Cl}}_{\text{2}} \text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$

${\text{mole fraction of CHCl}}_{\text{3}} \text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$

${\text{P}}_{\text{T}}={\text{X}}_{\text{C}{\text{H}}_{2}\text{C}{\text{l}}_{2}}\times {\text{vapour pressure}}_{\text{C}{\text{H}}_{2}\text{C}{\text{l}}_{2}}+ {\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}}\times {\text{vapour pressure}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}}$

$=415\times \frac{1}{2} +200\times \frac{1}{2} =307.5$

${\text{P}}_{\text{T}} { ( {\text{X}}_{\text{A}} ) }_{ \text{V}\text{P} } ={ ( {\text{X}}_{\text{A}} ) }_{ \text{L}\text{P} } \times { \text{ Vapour pressure of CHCl} }_{3}$

$307.5\times { ( {\text{X}}_{ \text{C}\text{H}\text{C}{\text{l}}_{3} } ) }_{ \text{V}\text{P} } =200\times \frac{1}{2}$

${\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}}=\frac{100}{307.5}=0.325$

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Solution

${\text{mole of CH}}_{2}\text{C}{\text{l}}_{2} \text{in liquid phase}=\frac{8.5}{85}=0.1$

${\text{mole of CHCl}}_{\text{3}} \text{in liquid phase}=\frac{11.95}{119.5}=0.1$

${\text{mole fraction of CH}}_{\text{2}}{\text{Cl}}_{\text{2}} \text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$

${\text{mole fraction of CHCl}}_{\text{3}} \text{in liquid phase}=\frac{0.1}{0.2}=\frac{1}{2}$

$=415\times \frac{1}{2} +200\times \frac{1}{2} =307.5$

${\text{P}}_{\text{T}} { ( {\text{X}}_{\text{A}} ) }_{ \text{V}\text{P} } ={ ( {\text{X}}_{\text{A}} ) }_{ \text{L}\text{P} } \times { \text{ Vapour pressure of CHCl} }_{3}$

$307.5\times { ( {\text{X}}_{ \text{C}\text{H}\text{C}{\text{l}}_{3} } ) }_{ \text{V}\text{P} } =200\times \frac{1}{2}$

${\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}}=\frac{100}{307.5}=0.325$

JEE Main 2017 — Chemistry Physical Chemistry

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JEE Main 2017 — Chemistry Physical Chemistry

Options

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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

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