The correct option for the value of vapour pressure of a solution at 45^oC with benzene to octane in molar ratio 3 : 2 is :
[At 45^oC vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

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336 mm of Hg

160 mm of Hg

350 mm of Hg

168 mm of Hg

answer is D.

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Detailed Solution

Applying Raoult’s law, $P_{total} = P_{A}^{0} \cdot X_{A} + P_{B}^{0} \cdot X_{B}$
$\begin{array}{l} X_{A} = \frac{3}{5}, X_{B} = \frac{2}{5} \\ P_{total} = 280 \times (\frac{3}{5}) + 420 \times (\frac{2}{5}) = 336 mm of Hg \end{array}$

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