When 25 g of a non-volatile solute is dissol | vedclass

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(C) Given:Weight of non-volatile solute,$w = 25 \ g$Weight of solvent,$W = 100 \ g$Lowering of vapour pressure,$p^{\circ} - p_s = 0.225 \ mm$Vapour pr

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$350$

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(C) Given:Weight of non-volatile solute,$w = 25 \ g$Weight of solvent,$W = 100 \ g$Lowering of vapour pressure,$p^{\circ} - p_s = 0.225 \ mm$Vapour pressure of pure solvent,$p^{\circ} = 17.5 \ mm$Molar mass of solvent $(H_2O)$,$M = 18 \ g/mol$Molar mass of solute,$m = ?$According to Raoult's law for lowering of vapour pressure:$\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{w 	imes M}{m 	imes W}$Substituting the values:$\frac{0.225}{17.5} = \frac{25 	imes 18}{m 	imes 100}$$m = \frac{25 	imes 18 	imes 17.5}{0.225 	imes 100}$$m = \frac{25 	imes 18 	imes 17.5}{22.5}$$m = 350 \ g/mol$

When $25 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of water,the vapour pressure is lowered by $0.225 \ mm$. If the vapour pressure of water at $20^{\circ}C$ is $17.5 \ mm$,what is the molecular weight of the solute?

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