Vapour pressure of chloroform (CHCI₃) and dichloromethane (CH₂Cl₂) at 25^o C are 200 mm of Hg and 41.5 mm of Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at the same temperature will be
(Molecular mass of CHCI₃ = 119.5 u and molecular mass of CH₂Cl₂ - 85 u)

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347.9 mm Hg

173.9 mm Hg

90.63 mm Hg

615.0 mm Hg

answer is D.

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Detailed Solution

Number of moles of CHCI₃,
$n_{A} = \frac{w}{M} = \frac{25 .5}{119 .5} = 0 .213$
Number of moles of CH₂Cl₂
$n_{B} = \frac{40}{85} = 0 .47$
Mole fraction of CHCI₃,
$x_{A} = \frac{n_{A}}{n_{A} + n_{B}} = \frac{0 .213}{0 .683} = 0 .31$
Mole fraction of CH₂Cl₂
$x_{B} = 1 - x_{A} = 1 - 0 .31 = 0 .69$
$\begin{array}{l} p_{total} = p_{A} x_{A} + p_{B} x_{B} \\ = 200 \times 0 .31 + 41 .5 \times 0 .693 \\ = 62 + 28 .63 = 90 .63 mmHg \end{array}$