## What is the boiling point of an aqueous solution of \(\mathrm{CaCl}_{2}\) with a molal concentration of 1.56 mol/kg?

Answer:

The boiling point of the solution is 108.705 °C

Explanation:

The formula for boiling point elevation \(\Delta T_{\mathrm{b}}\) is

where

\(i\) is the van’t Hoff \(i\) factor

\(K_{\mathrm{b}}\) is the molal boiling point elevation constant

\(b\) is the molality of the solution

In this problem,

\(i=3\), because 1 mol of \(\mathrm{CaCl}_{2}\) gives 3 mol of ions in solution.

\(K_{\mathrm{b}}=1.86{ }^{\circ} \mathrm{C} \cdot \mathrm{mol} \cdot \mathrm{kg}^{-1}\)

\(b=1.56 \mathrm{~mol} \cdot \mathrm{kg}^{-1}\)