QJEEAdv05Ch5eb0: JEE Advanced 2005 Chemistry - Class 12 JEEAdv05Ch5eb0 Solution: The elevation in boiling point of a solution of 13.44 g of CuCl …

We know, for ionic compounds, elevation in boiling point $\Delta$ T_b = k_b $\times$ molality (m) $\times$ i

Given, k_b = 0.52 K molal ^{$-$ 1}

Now, molality (m) = ${{mass\,of\,solute \div molar\,mass} \over {mass\,of\,solvent}} \times 1000$

or, $m = {{13.44/134.4} \over {1000}} \times 1000 = 0.1$ mol kg ^{$-$ 1}

We know, $\alpha = {{i - 1} \over {n - 1}}$

Now, 1 molecule of CuCl₂ on dissociation in solution produces 3 ions

CuCl₂(aq) $\to$ Cu²⁺(aq) + 2Cl ^$-$ (aq)

Hence, n = 3

If it is assumed that CuCl₂ undergoes complete dissociation in solution, then $\alpha$ = 1

$\therefore$ $1 = {{i - 1} \over {3 - 1}}$ or, i = 3

For a solution of an electrolyte, $\Delta$ T_b = i $\times$ k_b $\times$ m

$\therefore$ $\Delta$ T_b = 3 $\times$ 0.52 $\times$ 0.1 = 0.156 K

$\therefore$ Elevation of boiling point of the solution = 0.156 K