Question

A salt MA₂ ionises as
$\text{MA}_{2} \rightleftharpoons \text{M}^{2 +} + 2 \text{A}^{-}$
It was found that a given solution of the salt had the same freezing point as solution of glucose of twice the molality. The apparent degree of ionization of the salt is

• A. 0.25
• B. 0.33
• C. 0.50
• D. 0.67

Answer 1

Answer: (C)

$\text{MA}_{2} & \rightleftharpoons & \text{M}^{2 +} & + & 2\text{A}^{-} \\ 1-\text{α} & & \text{α} & & 2\text{α}$
$\text{i} = \frac{\text{Actual no. of particles in solution}}{\text{Number of particles taken}}$
$\text{i} = 1 + 2 \text{α}$
$\Delta \text{T}_{\text{} \text{} \text{f} }$ of $\text{MA}_{2} = \Delta \text{T}_{\text{} \text{} \text{f} }$ of glucose
$\text{i} \cdot \text{K}_{\text{f}} \cdot \text{m} = \text{K}_{\text{f}} \cdot 2 \text{m} \times \text{i}$
$1 + 2 \text{α} = 2 \Rightarrow 2 \text{α} = 1 \Rightarrow \text{α} = \text{0.5}$