JEE Main 2017 · 9 Apr · Shift 1 · Chemistry Q40

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ChemistrySingle correct+4 / -1Question 40 of 90

A solution is prepared by mixing 8.5 g of $\ce{CH2Cl2}$ and 11.95 g of $\ce{CHCl3}$ . If vapour pressure of $\ce{CH2Cl2}$ and $\ce{CHCl3}$ at 298 K are 415 and 200 mmHg respectively, the mole fraction of $\ce{CHCl3}$ in vapour form is :

(Molar mass of Cl = 35.5 g mol $^{-1}$ )

0.162

0.675

0.325

0.486

Answer and worked solution

Official answer: C

$\ce{CH2Cl2}$ : $\frac{8.5}{85}=0.1$ mol; $\ce{CHCl3}$ : $\frac{11.95}{119.5}=0.1$ mol. Each has mole fraction 0.5 in the liquid, so $p=207.5+100=307.5$ mmHg and
$y_{\ce{CHCl3}}=\frac{100}{307.5}=0.325.$

Topics:Raoult's law and vapour pressure

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JEE Main 2017 · 9 Apr · Shift 1, Chemistry question 40