In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionization is 0.3. Taking Kf for water as 1.85, the freezing point of the solution will be nearest to:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. –0.480°C
| \(HX \) | ⇌ | \(H^+\) | + | \( X^−\) |
| \(1\) | 0 | 0 | ||
| \(1 − 0.3\) | \(0.3\) | \(0.3\) |
Total number of moles after dissociation \(= 1 − 0.3 +0.3 + 0.3 \)
\(\frac{K_f\text{(observed)}}{K_f\text{(experimental)}} = \frac{\text{Number of moles after dissociation }}{\text{Number of moles before dissociation}}\)
or,\(\frac{K_f\text{(observed)}}{1.85} = \frac{1.3}{1}\)
or, \(K_f\text{(observed)}= 1.85 × 1.3 = 2.405\)
\(\Delta T_f = K_f × \text{ molality} = 2.405 ×0.2 = 0.4810\)
Freezing point of the solution \(= 0 − 0.481 = − 0.481^oC\)