# QuestionThe vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at  250C is 23.8 mm HgA.B.C.D.

As we learned

Expression of relative lowering of vapor pressure -

$\frac{\Delta P}{ P^{0}}= x_{solute}$

$x_{solute}= \frac{ n_{solute}}{ n_{solute}+n_{solvent}}$

- wherein

$\Delta P \: is \: lowering \: o\! f \: v.p.$

$P^{0}\rightarrow \: vapour\: pressure\: of \: pure\: solvent$

$x_{solute}\rightarrow \: mole\: fraction \: of \:non\: volatile\: solute$

Given the molecular mass of sucrose = 342

$Moles\: of \: sucrose =\frac{100}{342}=0.292mole$

$Moles\: of \: water \: N =\frac{1000}{18}=55.5mole\: and$

$Vapour\: pressure\: of \: pure \: water\: P^{0} =23.8mmHg$

$According \: to \: Raoult's \: law\; \frac{\Delta P}{P^{0}}=\frac{n}{n+N}\Rightarrow \frac{\Delta P}{23.8}=\frac{0.292}{0.292+55.5}$

$\Delta P=\frac{23.8\times 0.292}{55.792}=0.125mmHg$

Thus option B is correct.

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