Q38 / 150 · MHT CET 2026 (19 April Shift 2) · Chemistry
Calculate the molality of aqueous solution of electrolyte that freezes at $-0.93\,^\circ\text{C}$ if $K_f$ for water and van't Hoff factor respectively are $1.86\,\text{K kg mol}^{-1}$ and $1.25$. (Freezing point of water $= 0\,^\circ\text{C}$ )
- A$0.2$ m
- B$0.5$ m
- C$0.3$ m
- Correct answerD$0.4$ m
Solution
The depression in freezing point is given by the formula:
$\Delta T_f = i \times K_f \times m$
Given:
Freezing point of solution, $T_f = -0.93\,^\circ\text{C}$
Freezing point of water, $T_f^\circ = 0\,^\circ\text{C}$
$\Delta T_f = T_f^\circ - T_f = 0 - (-0.93) = 0.93\,^\circ\text{C}$
van't Hoff factor, $i = 1.25$
Cryoscopic constant, $K_f = 1.86\,\text{K kg mol}^{-1}$
Substituting the values into the formula:
$0.93 = 1.25 \times 1.86 \times m$
$m = \dfrac{0.93}{1.25 \times 1.86}$
$m = \dfrac{0.5}{1.25} = 0.4\,\text{m}$
Answer: $0.4$ m
$\Delta T_f = i \times K_f \times m$
Given:
Freezing point of solution, $T_f = -0.93\,^\circ\text{C}$
Freezing point of water, $T_f^\circ = 0\,^\circ\text{C}$
$\Delta T_f = T_f^\circ - T_f = 0 - (-0.93) = 0.93\,^\circ\text{C}$
van't Hoff factor, $i = 1.25$
Cryoscopic constant, $K_f = 1.86\,\text{K kg mol}^{-1}$
Substituting the values into the formula:
$0.93 = 1.25 \times 1.86 \times m$
$m = \dfrac{0.93}{1.25 \times 1.86}$
$m = \dfrac{0.5}{1.25} = 0.4\,\text{m}$
Answer: $0.4$ m
Hence, the correct answer is option: 'D'.