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Solutions question

2017 · 9 Apr · Shift 1 · Q18

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2017 · 9 Apr · Shift 1 · Q18

Solutions question

2017 · 9 Apr · Shift 1 · Q18

JEE MainChemistrySolutionsMCQ+4 / −1

A solution is prepared by mixing 8.5 g of

CH_2Cl_2

and 11.95 g of

CHCl_3

. If vapour pressure of

CH_2Cl_2

and

CHCl_3

at 298 K are 415 and 200 mmHg respectively, the mole fraction of

CHCl_3

in vapour form is : (Molar mass of Cl = 35.5 g mol−1)

A
0.162
B
0.675
C
0.325
D
0.486

View written solutionFree

Correct answer: C

Find molar masses

Given:

$CH_2Cl_2$ : $M = 12 + 2(1) + 2(35.5) = 85\,\text{g mol}^{-1}$
$CHCl_3$ : $M = 12 + 1 + 3(35.5) = 119.5\,\text{g mol}^{-1}$

Calculate moles of each component

For $CH_2Cl_2$ : $n_1 = \frac{8.5}{85} = 0.1\,\text{mol}$

For $CHCl_3$ : $n_2 = \frac{11.95}{119.5} = 0.1\,\text{mol}$

So total moles in liquid solution: $n_{\text{total}} = 0.1 + 0.1 = 0.2$

Calculate mole fractions in liquid phase

$x_{CH_2Cl_2} = \frac{0.1}{0.2} = 0.5$ $x_{CHCl_3} = \frac{0.1}{0.2} = 0.5$

Apply Raoult's law to get partial vapour pressures

For $CH_2Cl_2$ : $p_1 = x_1 p_1^0 = 0.5 \times 415 = 207.5\,\text{mmHg}$

For $CHCl_3$ : $p_2 = x_2 p_2^0 = 0.5 \times 200 = 100\,\text{mmHg}$

Find total vapour pressure

$P_{\text{total}} = 207.5 + 100 = 307.5\,\text{mmHg}$

Find mole fraction of $CHCl_3$ in vapour phase

If $y_{CHCl_3}$ is the vapour-phase mole fraction, then $y_{CHCl_3} = \frac{p_{CHCl_3}}{P_{\text{total}}} = \frac{100}{307.5}$

$y_{CHCl_3} \approx 0.325$

Match with options

Thus the mole fraction of $CHCl_3$ in vapour form is: $\boxed{0.325}$

So the correct option is C.

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