The elevation in boiling point (K) of a solution of 13.44 g of CuCl₂ in 1 kg ofwater using the following information will be (Molecular weight of CuCl₂ = 134.4 g and K_b = 0.52 K kg mol^-1)

(i) $\mathrm{i} = \frac{\mathrm{No}. \mathrm{of} \mathrm{particles} \mathrm{after} \mathrm{ionisation}}{\mathrm{No}. \mathrm{of} \mathrm{particles} \mathrm{before} \mathrm{ionisation} }$
$\text{ (ii) }{\mathrm{\Delta T}}_{\mathrm{b}}-\mathrm{i}\times {\mathrm{K}}_{\mathrm{b}}\times \mathrm{m}$
$\underset{\underset{(1-\mathrm{\alpha })}{1}}{{\mathrm{CuCl}}_2}⟶\underset{\underset{\mathrm{\alpha }}{0}}{{\mathrm{Cu}}^{2+}}+\underset{\underset{2\mathrm{\alpha }}{0}}{2{\mathrm{Cl}}^{-}}$
$\mathrm{i}=\frac{1+2\mathrm{\alpha }}{1},\mathrm{i}=1+2\mathrm{\alpha }$
Assuming 100%ionization
$\mathrm{So},\mathrm{i} = 1+2=3$
${\mathrm{\Delta T}}_{\mathrm{b}}=3\times 0.52\times 0.1=0.156\approx 0.16\left[\mathrm{m}=\frac{13.44}{134.4}=0.1\right]$