Question

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

• A. 0.70
• B. 0.30
• C. 0.125
• D. 0.07

Answer 1

Answer: (A)

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