Vapour pressure of $C C l_{4}$ at 25°C is 143mm of Hg0.5gm of a non-vola~le
solute (mo/. wt. = 65) is dissolved in $100 {mlCCl}_{4}$ . Find the vapour pressure
of the solution $(Density of {CCl}_{4} = 1.58 g / {cm}^{2})$

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94.39mm

199.34mm

$141.93 mm$

143.99 mm

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{P^{o} - P_{s}}{P^{o}} = \frac{w \times M}{m W} = 143 - \frac{0.5 \times 154}{65 \times 158} \times 143 \\ = 143 - 1.07 = 141.93 mm . \end{array}$

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