JEE Main 2025 — Solutions Question with Solution

Boiling Point Elevation Formula:

$\Delta T_b=i_1\cdot m_1\cdot K_b+i_2\cdot m_2\cdot K_b$

where:

$i_1$ and $i_2$ are the van't Hoff factors for ethylene glycol and glucose, respectively (both are 1 since they do not dissociate in solution).

$m_1$ and $m_2$ are the molalities of ethylene glycol and glucose, respectively.

$K_b$ is the ebullioscopic constant of water ($0.52{\text{K kg mol}}^{-1}$).

Calculate Molality:

Each solute has 2 moles dissolved in 500 grams of water ($0.5\text{kg}$):

$m_1=\frac{2\text{moles}}{0.5\text{kg}}=4{\text{mol kg}}^{-1}$

$m_2=\frac{2\text{moles}}{0.5\text{kg}}=4{\text{mol kg}}^{-1}$

Substitute into the Formula:

$\Delta T_b=1\cdot 4\cdot 0.52+1\cdot 4\cdot 0.52=4.16$

Determine Boiling Point of Solution:

The normal boiling point of water is $373.16\text{K}$.

Add the boiling point elevation to the normal boiling point:

$T_b(\text{solution})=373.16+4.16=377.3\text{K}$

Thus, the boiling point of the resulting solution is $377.3\text{K}$.