Q. Calculate the elevation in boiling point, for a solution of $\text{13} \text{.44} \, \text{g}$ of $CuCl_{2}$ in $1kg$ of water using the following information
(Molecular weight of $\text{CuCl}_{\text{2}} \, \text{=} \, \text{134} \text{.4}$ and $\text{K}_{\text{b}} \text{=} \, \text{0} \text{.52} \, \text{Kmolal}^{\text{1}}$ )

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Solution:

$\Delta T_{b}=i.K_{b}.m$
$CuCl_{2} \rightarrow & Cu^{2 +}+ & 2Cl \\ 1 & 0 & 0 \\ \left(\right.1-\alpha \left.\right) & \alpha & 2\alpha $
$i=1+2\alpha $
Assuming 100% ionization, $\alpha =1$
So, i = 3
$\Delta T_{b}=3\times 0.52\times 0.1=0.156\cong0.16$