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

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

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Solution:

According to Raoult's law, "Relative lowering of vapour pressure $=$ mole fraction of solute"
$\left(X_{B}\right)=0.0125$
$\because X_{B}=\frac{m \cdot M_{A}}{10000+m \cdot M_{A}}$
$\therefore 0.0125=\frac{m \times 18}{1000+m \times 18}$
$12.5+0.225 m=18 m$
$17.775 m=12.5$
$m=\frac{12.5}{17.775}=0.70$