On mixing, heptane and octane form an ideal solution. At $373 K$ , the vapour pressures of the two liquid components (heptane and octane) are $105 kPa$ and $45 kPa$
respectively. Vapour pressure of the solution obtained by mixing $25 g$ of heptane and $35$ g of octane will be (molar mass of heptane $= 100 g {mol}^{- 1}$ and of octane
$= 114 g {mol}^{- 1})$

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$36.1 kPa$

$144.5 kPa$

$96.2 kPa$

$72.0 kPa$

answer is B.

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Detailed Solution

If A= Heptane; B = octane

$\begin{array}{l} P_{S} = P_{A}^{o} X_{A} + P_{B}^{o} X_{B} \\ X_{A} = \frac{25 / 100}{\frac{25}{100} \times \frac{35}{114}} = \frac{0.25}{0.25 + 0.30} = \frac{0.25}{0.55} = \frac{5}{11} \\ Then X_{B} = \frac{6}{11} \\ P_{S} = 105 \times \frac{5}{11} + 45 \times \frac{6}{11} = 47.727 + 24.54 \approx 72 kPa \end{array}$