Question

5g of a nonvolatile compound (molar mass 50gmol−1) is added to 100 g of water . If the decrease of pressure is 11.31 Torr, the vapour pressure of water will be

Answer 1

As we learned  

 

Expression of relative lowering of vapour 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 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

 

Option 1)

1.25 mm Hg

Option 2)

0.125 mm Hg

Option 3)

1.15 mm Hg

Option 4)

00.12 mm Hg