The elevation in boiling point, when 13.44 g of freshly prepared are added to one kg of water, is (S — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Moles of $\text{CuCl}_2 = 13.44 / 134.4 = 0.1 \text{ mol}$. Molality $m = 0.1 \text{ mol} / 1 \text{ kg} = 0.1 \text{ m}$. Assuming complete dissociation, $\text{CuCl}_2 \rightarrow \text{Cu}^{2+} + 2\text{Cl}^-$, giving $i=3$. $\Delta T_b = i \times K_b \times m = 3 \times 0.52 \times 0.1 = 0.156 \approx 0.16^\circ\text{C}$. Therefore, correct answer is C.