JEE Chemistry 2025 Question with Solution

Given: Ethylene glycol and glucose are dissolved in $500 \, \text{g} = 0.5 \, \text{kg}$ of water, with $2$ moles each. Find: The boiling point of the resulting solution.

Both ethylene glycol and glucose are non-electrolytes, so for each solute, the van 't Hoff factor is $i = 1$.

Use the boiling point elevation relation:

$$\Delta T_b = i \cdot K_b \cdot m$$

For two non-electrolytes, the total molality is

$$m = \frac{2 + 2}{0.5} = \frac{4}{0.5} = 8 \; \text{mol kg}^{-1}$$

Now,

$$\Delta T_b = 1 \cdot 0.52 \cdot 8 = 4.16 \; \text{K}$$

The normal boiling point of water is taken as $373 \, \text{K}$. Therefore,

$$T_b = 373 + 4.16 = 377.16 \; \text{K}$$

Rounding to one decimal place gives $377.2 \, \text{K}$, which is closest to $377.3 \, \text{K}$.

Therefore, the correct option is B.

The alternate working shown is:

$$\Delta T_b = i_1 m_1 k_b + i_2 m_2 k_b$$ $$\Delta T_b = 1 \times \frac{2}{0.5} \times 0.52 + 1 \times \frac{2}{0.5} \times 0.52 = 4.16$$ $$(T_b)_{\text{solution}} = 373.16 + 4.16 = 377.3 \, \text{K}$$

This supports option B.

Given: Two non-electrolyte solutes, each of $2$ moles, in $0.5 \, \text{kg}$ water. Find: Boiling point of the solution.

Since both solutes are non-electrolytes, their effects on boiling point elevation add directly. So,

$$\Delta T_b = i_1 m_1 K_b + i_2 m_2 K_b$$ $$= 1 \times \frac{2}{0.5} \times 0.52 + 1 \times \frac{2}{0.5} \times 0.52 = 4.16 \; \text{K}$$

Hence the boiling point is approximately

$$373.16 + 4.16 = 377.3 \; \text{K}$$

So the correct option is B.