2.5 g of a non-volatile,non-electrolyte is d | vedclass

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(D) Given: $\Delta T_b = 2 \ K$,$K_b = 0.52 \ K \ kg \ mol^{-1}$,$W_{	ext{solvent}} = 100 \ g$,$M_{	ext{solvent}} = 18 \ g \ mol^{-1}$,$P^{\circ} =

Vedclass · Accepted Answer

$707$

Vedclass · Answer

(D) Given: $\Delta T_b = 2 \ K$,$K_b = 0.52 \ K \ kg \ mol^{-1}$,$W_{	ext{solvent}} = 100 \ g$,$M_{	ext{solvent}} = 18 \ g \ mol^{-1}$,$P^{\circ} = 760 \ mm \ Hg$.Step $1$: Calculate molality $(m)$ using $\Delta T_b = K_b 	imes m$.$2 = 0.52 	imes m \implies m = \frac{2}{0.52} \approx 3.846 \ mol \ kg^{-1}$.Step $2$: Use Raoult's law for dilute solutions: $\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_{	ext{solute}}}{n_{	ext{solvent}}}$.Since $m = \frac{n_{	ext{solute}}}{W_{	ext{solvent}}(kg)}$,we have $n_{	ext{solute}} = m 	imes \frac{W_{	ext{solvent}}}{1000}$.$n_{	ext{solvent}} = \frac{100}{18} = 5.556 \ mol$.Step $3$: Calculate relative lowering of vapour pressure.$\frac{\Delta P}{P^{\circ}} = \frac{m 	imes W_{	ext{solvent}}}{1000} 	imes \frac{1}{n_{	ext{solvent}}} = \frac{m 	imes M_{	ext{solvent}}}{1000}$.$\Delta P = 760 	imes \frac{3.846 	imes 18}{1000} = 760 	imes 0.069228 = 52.613 \ mm \ Hg$.Step $4$: Calculate vapour pressure of solution $(P_s)$.$P_s = P^{\circ} - \Delta P = 760 - 52.613 = 707.387 \ mm \ Hg$.Rounding to the nearest integer,we get $707 \ mm \ Hg$.

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