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(A) The loss in weight of pure water is proportional to the vapour pressure of pure water $(P^0)$,and the loss in weight of the solution is proportion

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The loss in weight of pure water is proportional to the vapour pressure of pure water $(P^0)$,and the loss in weight of the solution is proportional to the vapour pressure of the solution $(P_s)$.Let the loss in weight of pure water be $w_1 = 1.5 + 2 = 3.5 \ g$.Let the loss in weight of the solution be $w_2 = 2 \ g$.According to the relative lowering of vapour pressure formula: $\frac{P^0 - P_s}{P^0} = \frac{w_1 - w_2}{w_1} = \frac{n}{n + N}$.Here,$n$ is the number of moles of solute and $N$ is the number of moles of solvent (water).$N = \frac{360 \ g}{18 \ g/mol} = 20 \ mol$.Substituting the values: $\frac{3.5 - 2}{3.5} = \frac{1.5}{3.5} = \frac{3}{7} = \frac{n}{n + 20}$.$3(n + 20) = 7n$ $\Rightarrow 3n + 60 = 7n$ $\Rightarrow 4n = 60$ $\Rightarrow n = 15 \ mol$.Since $n = \frac{	ext{mass of solute}}{	ext{molar mass of solute}} = \frac{600}{M} = 15$.$M = \frac{600}{15} = 40 \ g/mol$.

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