JEE Main & Advanced Chemistry Solutions / विलयन Sample Paper Topic Test - Solutions

The relative lowering of vapour pressure of aqueous solution containing a non-volatile solute is 0.0125. The molality of the solution is

A) 0.70

B) 0.50

C) 0.80

D) 0.40

Correct Answer: A

Solution :

[a] \[\frac{p{}^\circ -p}{p{}^\circ }={{X}_{1}}=0.0125=\] mole fraction of solute

Mole fraction and molarity are related as

\[\frac{1}{X}-1=\frac{1000}{M\times m}\]

\[\frac{1}{0.0125}-1=\frac{1000}{18\times m}\]

\[m=0.70\]