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Relative Lowering of Vapour Pressure

At 20∘C The v...

Question

At $20^{\circ} C$ The vapour pressure of pure ether is $442 m m H g$ . When $6.1 g$ of a non volatile substance is dissolved in $50 g$ of ether (molecular weight 74). The vapour pressure of solution falls to $410 m m H g$ .
What is the molecular weight of the substance?
Assume the solution to be non-dilute.

A

74.2 g

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B

33.9 g

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C

115.6 g

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D

50.5 g

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Solution

The correct option is C $115.6 g$
For a non-dilute solution:

$\Rightarrow \frac{n_{s o l u t e}}{n_{s o l v e n t}} = \frac{P_{s o l v e n t}^{\circ} - P_{s o l u t i o n}}{P_{s o l u t i o n}} = \frac{442 - 410}{410}$

$\Rightarrow \frac{\frac{6.1}{M_{s o l u t e}}}{\frac{50}{74}} = \frac{32}{410}$

$M_{s o l u t e} = \frac{6.1 \times 74 \times 410}{50 \times 32} = 115.67 g$

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Relative Lowering of Vapour Pressure

Standard XII Chemistry

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