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

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

KEAMKEAM 2007Solutions

Solution:

Relative lowering of vapour pressure = mole fraction of solute (Raoults law)
$ \frac{P-{{P}_{S}}}{P}={{X}_{2}} $
$ \frac{P-{{P}_{S}}}{P}=\frac{wM}{mW} $
where w = wt. of solute M = mol. wt. of solvent m = mol. wt of solute W = wt. of solvent
$ 0.0125=\frac{wM}{mW} $
Or, $ \frac{w}{mW}=\frac{0.0125}{18}=0.00070 $
Hence, molality $ =\frac{w}{mW}\times 1000 $
$ =0.0007\times 1000=0.70 $