Q115 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Chemistry
A solution containing 1.2 g nonvolatile solute in 100 g. of a solvent has a boiling point elevation 1.2 K. Calculate the molar mass of solute if molal elevation constant of solvent is $6.5\ \text{K kg mol}^{-1}$.
- A$60\ \text{g mol}^{-1}$
- Correct answerB$65\ \text{g mol}^{-1}$
- C$70\ \text{g mol}^{-1}$
- D$75\ \text{g mol}^{-1}$
Solution
The elevation in boiling point is given by the formula:
$\Delta T_b = K_b \times m$
where $m$ is the molality of the solution, which can be written as:
$m = \dfrac{w_2 \times 1000}{M_2 \times w_1}$
Substituting the given values $w_2 = 1.2\ \text{g}$, $w_1 = 100\ \text{g}$, $\Delta T_b = 1.2\ \text{K}$, and $K_b = 6.5\ \text{K kg mol}^{-1}$:
$1.2 = 6.5 \times \dfrac{1.2 \times 1000}{M_2 \times 100}$
$1.2 = 6.5 \times \dfrac{12}{M_2}$
$M_2 = \dfrac{6.5 \times 12}{1.2}$
$M_2 = 65\ \text{g mol}^{-1}$
Answer: $65\ \text{g mol}^{-1}$
$\Delta T_b = K_b \times m$
where $m$ is the molality of the solution, which can be written as:
$m = \dfrac{w_2 \times 1000}{M_2 \times w_1}$
Substituting the given values $w_2 = 1.2\ \text{g}$, $w_1 = 100\ \text{g}$, $\Delta T_b = 1.2\ \text{K}$, and $K_b = 6.5\ \text{K kg mol}^{-1}$:
$1.2 = 6.5 \times \dfrac{1.2 \times 1000}{M_2 \times 100}$
$1.2 = 6.5 \times \dfrac{12}{M_2}$
$M_2 = \dfrac{6.5 \times 12}{1.2}$
$M_2 = 65\ \text{g mol}^{-1}$
Answer: $65\ \text{g mol}^{-1}$
Hence, the correct answer is option: 'B'.