The mass of a non-volatile, non-electrolyte solute ( $m o l a r m a s s = 50 g m o l^{- 1}$ ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75% is

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37.5 g

150 g

50 g

75 g

answer is B.

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Detailed Solution

Given data:
It is given to identify the mass of a non-volatile, non-electrolyte solute.
Explanation:
$T h e m o l a r m a s s o f o c \tan e i s 114 g / m o l . \frac{Δ P}{P} = \frac{\frac{W_{2}}{M_{2}}}{\frac{W_{2}}{M_{2}} + \frac{W_{1}}{M_{1}}} \frac{75}{100} = \frac{\frac{W_{2}}{50 g / mol}}{\frac{W_{2}}{50 g / mol} + \frac{114 g}{114 g / mol}} 0.75 = \frac{\frac{W_{2}}{50}}{\frac{W_{2}}{50} + 1} \begin{array}{l} \frac{W_{2}}{50} + 1 = \frac{W_{2}}{50 \times 0.75} \\ W_{2} = 150 g \end{array}$