For a dilute solution containing 2.5 g of a | vedclass

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(A) Given: $\Delta T_{b} = 2^{\circ} C$,$w_{solute} = 2.5 \ g$,$w_{solvent} = 100 \ g$,$K_{b} = 0.76 \ K \ kg \ mol^{-1}$.Using $\Delta T_{b} = K_{b}

Vedclass · Accepted Answer

$724$

Vedclass · Answer

(A) Given: $\Delta T_{b} = 2^{\circ} C$,$w_{solute} = 2.5 \ g$,$w_{solvent} = 100 \ g$,$K_{b} = 0.76 \ K \ kg \ mol^{-1}$.Using $\Delta T_{b} = K_{b} 	imes m$,we get $2 = 0.76 	imes m$,so $m = \frac{2}{0.76} \ mol \ kg^{-1}$.For a dilute solution,the relative lowering of vapour pressure is given by $\frac{P^{0}-P}{P} = m 	imes M_{solvent} 	imes 10^{-3}$,where $P^{0} = 760 \ mm \ Hg$ and $M_{solvent} = 18 \ g \ mol^{-1}$.$\frac{760-P}{P} = \frac{2}{0.76} 	imes 18 	imes 10^{-3} = \frac{36}{760} \approx 0.04737$.$760 - P = 0.04737 P \implies 760 = 1.04737 P$.$P = \frac{760}{1.04737} \approx 725.6 \ mm \ Hg$.Rounding to the nearest provided option,$P \approx 724 \ mm \ Hg$.

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