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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

  • Option 1)

    1.25 mm Hg

  • Option 2)

    0.125 mm Hg

  • Option 3)

    1.15 mm Hg

  • Option 4)

    00.12 mm Hg

 

Answers (1)

best_answer

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

Posted by

satyajeet

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