Question

The ratio of lowering of vapour pressure of $0.1M$ aqueous solutions of $KCl$, $CuS{O}_4$ & $N{a}_{2}S{O}_4$ is:

• A. $1:2:3$
• B. $2:2:3$
• C. $1:1:1$
• D. $3:2:1$

Answer 1

Answer: (B)

$2:2:3$

As per Raoult's law,
$\frac{p_{final}-p^{\circ }}{p^{\circ }}=i\times \frac{n_{solute}}{n_{solvent}}$
Ratio of relative lowering is equivalent to ratio of van't hoff factor.
Van't hoff factor for $KCl$ is 2.
Van't hoff factor for $CuS{O}_4=2$
Van't hoff factor for $N{a}_{2}S{O}_4=3$