The ratio of lowering of vapour pressure of 0.1M aqueous solutions of $\mathrm{KCl}, {\mathrm{CuSO}}_4 \mathrm{and} {\mathrm{Na}}_2{\mathrm{SO}}_4 \mathrm{is}$

0.1M KCl $\to$ 0.2 moles of ion $\left(\mathrm{KCl}\to {\mathrm{K}}^{+}+{\mathrm{Cl}}^{-}\right)$

$0.1\mathrm{M} {\mathrm{CuSO}}_4\to 0.2 \mathrm{moles} \mathrm{of}\hspace{0.17em}\mathrm{ion} \left({\mathrm{CuSO}}_4\to {\mathrm{Cu}}^{2+}+{\mathrm{SO}}_4^{2-}\right)$$0.1\mathrm{M}\to 0.3\mathrm{moles}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\mathrm{ion} \left({\mathrm{Na}}_2{\mathrm{SO}}_4\to 2{\mathrm{Na}}^{+}+{\mathrm{SO}}_4^{2-}\right)$Ratio of lowering of vapour pressure depends upon no. of moles of ions.

=0.2:0.2:0.3

$\Rightarrow 2:2:3$