Vapour pressure of a solution of 5 g of non- | vedclass

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(C) According to Raoult's law for dilute solutions,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^o -

Vedclass · Accepted Answer

$180$

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(C) According to Raoult's law for dilute solutions,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1} = \frac{W_2 / M_2}{W_1 / M_1}$.Given: $P^o = 3000 \ N/m^2$,$P_s = 2985 \ N/m^2$,$W_2 = 5 \ g$,$W_1 = 100 \ g$,$M_1 = 18 \ g/mol$ (for water).Substituting the values: $\frac{3000 - 2985}{3000} = \frac{5 / M_2}{100 / 18}$.$\frac{15}{3000} = \frac{5 	imes 18}{100 	imes M_2}$.$0.005 = \frac{90}{100 	imes M_2}$.$0.005 = \frac{0.9}{M_2}$.$M_2 = \frac{0.9}{0.005} = 180 \ g/mol$.

Vapour pressure of a solution of $5 \ g$ of non-electrolyte in $100 \ g$ of water at a particular temperature is $2985 \ N/m^2$. The vapour pressure of pure water is $3000 \ N/m^2$. The molecular weight of the solute is

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