Q35 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
A solution of non volatile solute has boiling point elevation $0.70$ K if $K_b$ for the solvent is $2.44\ \text{K kg mol}^{-1}$. What is the molality of the solution?
- A$3.48$ m
- B$2.86$ m
- Correct answerC$0.286$ m
- D$0.0186$ m
Solution
The elevation in boiling point is given by the formula:
$\Delta T_b = K_b \times m$
where $\Delta T_b$ is the boiling point elevation, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution.
Given:
$\Delta T_b = 0.70$ K
$K_b = 2.44$ K kg mol$^{-1}$
Substituting the given values into the formula:
$0.70 = 2.44 \times m$
$m = \dfrac{0.70}{2.44}$
$m \approx 0.286$ m
Answer: $0.286$ m
$\Delta T_b = K_b \times m$
where $\Delta T_b$ is the boiling point elevation, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution.
Given:
$\Delta T_b = 0.70$ K
$K_b = 2.44$ K kg mol$^{-1}$
Substituting the given values into the formula:
$0.70 = 2.44 \times m$
$m = \dfrac{0.70}{2.44}$
$m \approx 0.286$ m
Answer: $0.286$ m
Hence, the correct answer is option: 'C'.