JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600\mathrm{mm}\mathrm{Hg}$ respectively at $298 K.$ On mixing the two liquids, the sum of their volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is $0.5$ in the mixture. The vapour pressure of the final solution, the mole fractions of components $A$ and $B$ in the vapour phase, respectively are

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

A
$500 mm Hg, 0.5, 0.5$
B
$450 mm Hg, 0.4, 0.6$
C
$450 mm Hg, 0.5, 0.5$
D
$500 mm Hg, 0.4, 0.6$

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Solution

Raoult’s law for ideal solution,

${P}_{Total}={P}_{A}+ {P}_{B}$

${P}_{Total}={x}_{A}\times {P}_{A}^{\circ }+ {x}_{B}\times {P}_{B}^{\circ }$

${P}_{Total}=\frac{1}{2}\times 400+\frac{1}{2}\times 600=500 mm of Hg$

Dalton’s law,

${y}_{A}\times {P}_{Total}={P}_{A}$

${y}_{A}=\frac{{x}_{A}\times {P}_{A}^{0}}{{P}_{Total}}=\frac{1}{2}\times \frac{400}{500}=0.4$

${y}_{B}=1-0.4=0.6$

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Solution

Raoult’s law for ideal solution,

${P}_{Total}={P}_{A}+ {P}_{B}$

${P}_{Total}={x}_{A}\times {P}_{A}^{\circ }+ {x}_{B}\times {P}_{B}^{\circ }$

${P}_{Total}=\frac{1}{2}\times 400+\frac{1}{2}\times 600=500 mm of Hg$

Dalton’s law,

${y}_{A}\times {P}_{Total}={P}_{A}$

${y}_{A}=\frac{{x}_{A}\times {P}_{A}^{0}}{{P}_{Total}}=\frac{1}{2}\times \frac{400}{500}=0.4$

${y}_{B}=1-0.4=0.6$

JEE Main 2019 — Chemistry Physical Chemistry

Options

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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