A solution is prepared by mixing 8.5 g of \(\mathrm{CH_2Cl_2}\) and 11.95 g of \(\mathrm{CHCl_3}\). If vapour pressure of \(\mathrm{CH_2Cl_2}\) and \(\mathrm{CHCl_3}\) at 298 K are 415 and 200 mmHg respectively, the mole fraction of \(\mathrm{CHCl_3}\) in vapour form is :
(Molar mass of Cl = 35.5 g \(\mathrm{mol^{-1}}\))
(Molar mass of Cl = 35.5 g \(\mathrm{mol^{-1}}\))
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- 0.162
- 0.675
- 0.325
- 0.486
The Correct Option is C
Solution and Explanation
Molar mass of \(\mathrm{CHCl_3} = 12 + 1 + 106.5 = 119.5\) g/mol, so \(n = 11.95/119.5 = 0.1\) mol.
Liquid mole fractions: each is 0.5.
\(p_{CH_2Cl_2} = 0.5\times415 = 207.5\) mmHg; \(p_{CHCl_3} = 0.5\times200 = 100\) mmHg.
\(p_{total} = 307.5\) mmHg.
Vapour mole fraction of \(\mathrm{CHCl_3}\): \(y = \frac{100}{307.5} = 0.325\).
Answer: (3) 0.325
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