Air was drawn through a solution containing 40 g of solute (non-volatile and non-electrolyte) in 100

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Air was drawn through a solution containing 40 g of solute (non-volatile and non-electrolyte) in 100 g of water and then through water. The loss of mass of water was 0.05 g and the total mass of water absorbed in sulphuric acid tube was 2.05 g. The molecular mass of the solute is

Accepted Answer

Loss is mass of solvent, w1 $\propto$ (P° – P) and gain is mass of absorbent, w2 $\propto$ P°$	herefore\frac{w_{1}}{w_{2}} = \frac{P^{o} - P}{P^{o}} = X_{1} \Rightarrow \frac{0.05}{2.05} = \frac{40/M}{\frac{40}{M} + \frac{100}{18}} \Rightarrow M = 288$

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