Question

Calculate elevation of boiling point when 4 g of MgSO\(_4\) is dissolved in 100 g water. (Molar mass = 120 g/mol), \( K_b = 0.52\ \text{K kg mol}^{-1} \), assume complete ionisation.

Hint:

Always multiply molality with van’t Hoff factor when electrolyte dissociates.

Answer 1

Step 1: Molality \[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{4/120}{0.1} = \frac{1}{3} \] Step 2: van’t Hoff factor (i) MgSO\(_4\) dissociates completely → \( \text{Mg}^{2+} + \text{SO}_4^{2-} \Rightarrow i = 2 \) Step 3: Boiling point elevation \[ \Delta T_b = i \cdot K_b \cdot m = 2 \cdot 0.52 \cdot \frac{1}{3} = 0.346\ \text{K} \] \[ \boxed{\Delta T_b = 0.35\ \text{K (approx)}} \]