Q1 / 150 · MHT CET 2026 (17 April Shift 1) · Chemistry
Calculate the molar mass of nonvolatile solute when 4 g of it is dissolved in 100 g solvent that boils at 319.4 K [$K_b$ = 2.4 K kg $\text{mol}^{-1}$; B.P of pure solvent = 319 K]
- A$220\ \text{gmol}^{-1}$
- B$230\ \text{gmol}^{-1}$
- Correct answerC$240\ \text{gmol}^{-1}$
- D$250\ \text{gmol}^{-1}$
Solution
Elevation in boiling point is given by $\Delta T_b = T_b - T_b^0$
$\Delta T_b = 319.4 - 319 = 0.4$ K
Using the formula for elevation in boiling point:
$\Delta T_b = K_b \times m = K_b \times \dfrac{w_2 \times 1000}{M_2 \times w_1}$
Substituting the given values:
$0.4 = 2.4 \times \dfrac{4 \times 1000}{M_2 \times 100}$
$0.4 = \dfrac{96}{M_2}$
$M_2 = \dfrac{96}{0.4} = 240$ g mol$^{-1}$
Answer: $240\ \text{gmol}^{-1}$
$\Delta T_b = 319.4 - 319 = 0.4$ K
Using the formula for elevation in boiling point:
$\Delta T_b = K_b \times m = K_b \times \dfrac{w_2 \times 1000}{M_2 \times w_1}$
Substituting the given values:
$0.4 = 2.4 \times \dfrac{4 \times 1000}{M_2 \times 100}$
$0.4 = \dfrac{96}{M_2}$
$M_2 = \dfrac{96}{0.4} = 240$ g mol$^{-1}$
Answer: $240\ \text{gmol}^{-1}$
Hence, the correct answer is option: 'C'.