If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80%, the molar mass (in g ${mol}^{- 1}$ ) of the solute is [Given that molar mass of n-octane is 114 g ${mol}^{- 1}$ ]

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$20$

$40$

$60$

$80$

answer is B.

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Detailed Solution

Assuming dilute solution

$\begin{array}{l} \frac{ΔP}{P_{A}^{0}} = \frac{n_{B}}{n_{A}} = \frac{w_{B}}{m_{B}} \cdot \frac{m_{A}}{w_{A}} \\ \frac{20}{100} = \frac{8}{m_{B}} \cdot \frac{114}{114} \\ m_{B} = \frac{8 \times 100}{20} = 40 {gmol}^{- 1} \end{array}$

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