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Q.

A dry air is passed through the solution, containing the 10 gm of solute and 90 gm of water and then it pass through pure water. There is the depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. Calculate the molecular weight of solute:

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a

180

b

100

c

50

d

25

answer is C.

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Detailed Solution

Lowering in weight of solution $\propto$ solution pressure
Lowering in weight of solvent $\propto P^{0} - P_{s}$
( $∵$ P⁰ = vapour pressure of pure solvent)
$\begin{array}{l} \frac{p^{0} - p_{s}}{p_{s}} = \frac{Lowering in weight of solvent}{Lowering in weight of solution} \\ \frac{p^{0} - p_{s}}{p_{s}} = \frac{w \times M}{m \times W}; \frac{0 .05}{2 .5} = \frac{10 \times 18}{90 \times m} \\ \Rightarrow m = \frac{2 \times 2 .5}{0 .05} = \frac{2 \times 250}{5} = 100 \end{array}$

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