A solution containing 30 , g of non-volatile | vedclass

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(C) Let $M$ be the molecular weight of the solute and $P^0$ be the vapour pressure of pure water at $25 \, ^oC$.According to Raoult's law for dilute s

Vedclass · Accepted Answer

$67.83$

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(C) Let $M$ be the molecular weight of the solute and $P^0$ be the vapour pressure of pure water at $25 \, ^oC$.According to Raoult's law for dilute solutions,$\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{w_2 	imes M_1}{M_2 	imes w_1}$.For the first case: $\frac{P^0 - 21.85}{21.85} = \frac{30 	imes 18}{M 	imes 90} = \frac{6}{M} \implies P^0 - 21.85 = \frac{131.1}{M} \dots (i)$.For the second case,after adding $18 \, g$ of water,the new mass of water is $90 + 18 = 108 \, g$: $\frac{P^0 - 22.15}{22.15} = \frac{30 	imes 18}{M 	imes 108} = \frac{5}{M} \implies P^0 - 22.15 = \frac{110.75}{M} \dots (ii)$.Subtracting equation $(ii)$ from $(i)$: $(P^0 - 21.85) - (P^0 - 22.15) = \frac{131.1}{M} - \frac{110.75}{M}$.$0.30 = \frac{20.35}{M}$.$M = \frac{20.35}{0.30} = 67.83 \, g \, mol^{-1}$.

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