A solution containing $$2.5 \times 10^{-3}$$ kg of a solute dissolved in $$75 \times 10^{-3}$$ kg of water boils at $$373.535$$ K. The molar mass of the solute is ______ mol$$^{-1}$$. [nearest integer] (Given : $$K_b(H_2O) = 0.52$$ K kg mol$$^{-1}$$ and boiling point of water $$= 373.15$$ K)

A solution containing $$2.5 \times 10^{-3}$$ kg of a solute dissolved in $$75 \times 10^{-3}$$ kg of water boils at $$373.535$$ K.

Mass of solute:

$$w_2 = 2.5 \times 10^{-3}\,\text{kg} = 2.5\,\text{g}$$

Mass of solvent:

$$w_1 = 75 \times 10^{-3}\,\text{kg} = 0.075\,\text{kg}$$

Boiling point elevation:

$$\Delta T_b = 373.535 - 373.15$$

$$= 0.385\,K$$

Using:

$$\Delta T_b = K_b m$$

$$m = \frac{\Delta T_b}{K_b}$$

$$m = \frac{0.385}{0.52}$$

$$= 0.7404\,\text{mol kg}^{-1}$$

Now,

$$m = \frac{w_2/M}{w_1}$$

$$0.7404 = \frac{2.5/M}{0.075}$$

$$0.7404 \times 0.075 = \frac{2.5}{M}$$

$$0.05553 = \frac{2.5}{M}$$

$$M = \frac{2.5}{0.05553}$$

$$M \approx 45\,\text{g mol}^{-1}$$

Final Answer:

$${45\,\text{g mol}^{-1}}$$