2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is :
(Given : Ebullioscopic constant of water = 0.52 K kg mol$$^{-1}$$)

The normal boiling point of pure water is $$373 \text{ K}$$.

Total moles of solute:
Ethylene glycol $$= 2 \text{ mol}$$
Glucose $$= 2 \text{ mol}$$
Therefore, $$n_{\text{total}} = 2 + 2 = 4 \text{ mol}$$.

Mass of solvent (water) $$= 500 \text{ g} = 0.5 \text{ kg}$$.

Molality is defined as $$m = \frac{n_{\text{solute}}}{\text{mass of solvent in kg}}$$. Hence
$$m = \frac{4}{0.5} = 8 \text{ mol kg}^{-1}$$ $$-(1)$$

The elevation in boiling point is given by the formula
$$\Delta T_b = i \, K_b \, m$$ $$-(2)$$
For non-electrolytes like ethylene glycol and glucose, the van’t Hoff factor $$i = 1$$.
Given $$K_b = 0.52 \text{ K kg mol}^{-1}$$.

Substituting values into $$(2)$$:
$$\Delta T_b = 1 \times 0.52 \times 8 = 4.16 \text{ K}$$.

Boiling point of the solution:
$$T_{\text{boil}} = 373 \text{ K} + 4.16 \text{ K} = 377.16 \text{ K}$$.

Rounded to one decimal place, $$T_{\text{boil}} \approx 377.3 \text{ K}$$.

Hence, the boiling point of the solution is $$377.3 \text{ K}$$ (Option B).