Explanation

(a)
$CuCl_{2}is\ an\ electrolyte\ which\ ionise\ in\ solution\$
$as\ follows.\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ CuCl_{2}\ \ \ \ \ \ \rightleftharpoons \ \ Cu^{2 +}\ \ \ + \ \ \ \ \ \ 2Cl^{-}$
At t=0 1 mole 0 0
After ionisation$(1 - \alpha)mole\ \ \ \alpha mole\ \ \ \ \ \ \ 2\alpha mole$
Thus, number of particles after ionisation
=1-$\alpha$+$\ \alpha + 2\alpha = 1 + 2\alpha$
$\therefore$ van,t Hoff factor (i)
$= \frac{number\ of\ particloes\ after\ ionisation\ }{number\ of\ particles\ \ before\ ionisation\ }$
$or\ \ \ \ \ \ \ (i) = \frac{1 + 2\alpha}{1}(On\ 100\ \%\ ionisation\ \alpha = 1)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1 + 2 \times 1}{1} = 3$
The elevation in boiling point (when colligative property is abnormal)
$\mathrm{\Delta}T^{b} = i \times k_{b} \times m$
$m \rightarrow$molality of solution
Molality of $CuCl_{2}$ solution
$\frac{\frac{weight\ of\ CuCl_{2}in\ gram}{mol.weight\ of\ CuCl_{2}}}{weight\ of\ water\ (solvent)in\ kg} = \frac{\frac{13.44}{134.4}}{1} = 0.1\ m$
Thus, $\mathrm{\Delta}T_{b} = 3 \times 0.52 \times 0.1 = 0.156 \approx 0.16℃$