Question

The vapour pressure of a dilute aqueous solution of glucose is $750$ mm Hg at $373$ K.

Calculate (i) molality, (ii) mole fraction of the solute.

• A. $0.73,0.0132$
• B. $0.078,0.02$
• C. $0.0729,0.0187$
• D. $0.0779,0.3$

Answer 1

The correct option is A $0.73,0.0132$

$P_0=$ Vapour pressure of water at $373K=760$ mm Hg

Using Raoult's law in the following form,

$\frac{P_0-P_s}{P_0}=\frac{w\times M}{W\times m}$

or $\frac{760-750}{760}=\frac{w}{W\times m}\times 18$

or $\frac{w}{W\times m}=\frac{10}{760\times 18}$

Molality $=\frac{w}{W\times m}\times 1000=\frac{10\times 1000}{760\times 18}=0.73m$

$P_s=$ Mole fraction of solvent $\times P_0$;

Mole fraction of solvent $=\frac{750}{760}$

So, mole fraction of solute $=(1-\frac{750}{760})=0.0132$