The vapour pressure of pure benzene and at a certain temperature is $640 m m$ $H g$ . A non-volatile, non-electrolyte solid weighing $2.175 g$ is added to $39.0 g$ of benzene. The vapour pressure of the solution is $600 m m$ $H g$ .
What is the molecular weight of the solid substance?

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Solution

Given that

$P^{o} = 640 m m H g, P_{S} = 600 m m H g$

$W_{2} = 2.175 g, W_{1} = 39.0 g$

$M w_{1} = 78$

$\frac{P^{o} - P_{S}}{P^{o}} = \frac{W_{2} \times M w_{1}}{M w_{2} \times W_{1}}$

$∴ \frac{640 - 600}{640} = \frac{2.175 \times 78}{M w_{2} \times 39}$

$∴ M w_{2} = 69.6 \approx 70$

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