Question

The 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 Hg

A.

1.25 mm Hg

B.

0.125 mm Hg

C.

1.15 mm Hg

D.

00.12 mm Hg

Answers (1)

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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