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# Engineering

Question

An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is

A.

0.02

B.

1

C.

20

D.

180

Answers (1)
P Pankaj Sanodiya

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

 

 \frac{P^{0}-P_{s}}{P^{0}}=\frac{18\times 18}{180\times 90}=0.02

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