$(M.wt.$ of $CuCl_2 =134.4 $ અને $K_b = 0.52\, , molal^{-1})$
$CuCl_2$ $\to $ $Cu^{2+}$ + $2Cl^-$
$1$ $0$ $0$
$(1-\alpha$) $\alpha$ 2$\alpha$
$i = 1 + 2$ $\alpha$
Assuming $ 100\%$ ionization
So, $i = 3$
$T_b = 3$ $\times$ $0.52$ $\times$ $0.1 = 0.156 $ $\approx$ $0.16$
[ઉપયોગ : ${R}=0.083\, {~L}\, bar \,{mol}^{-1} \,{~K}^{-1}$ ]
આપેલું છેઃ પરમાણુ દળ $:c =12$$H=1,$$CI= 35.5$,$CHCl_3$ની ઘનતા$= 1.49\,g\,cm^3$