Chemistry Q54 – JEE-MAIN 2025

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}

)

Clue 1 of 3: Colligative Properties

The boiling point of a solvent increases when a non-volatile solute is added, a phenomenon known as boiling point elevation.

🥷

Ninja Strategy•Boiling Point Elevation Principle

Recognize that adding a non-volatile solute increases the boiling point, immediately eliminating options below the normal boiling point of water. Then, estimate the magnitude of elevation to narrow down the remaining options.

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Step 1: Calculate the total molality of the solution✦ Active

Both ethylene glycol and glucose are non-electrolytes. The total moles of solute are the sum of moles of each component:

Total moles of solute = Moles of ethylene glycol + Moles of glucose = 2 mol + 2 mol = 4 mol

The mass of the solvent (water) is given as $500 g$ , which is $0.5 kg$ . Molality ( $m$ ) is calculated as:

m = \frac{Total moles of solute}{Mass of solvent in kg} = \frac{4 mol}{0.5 kg} = 8 {mol kg}^{- 1}

💡 Teacher's Secret Hint

Ensure the mass of the solvent is converted to kilograms for molality calculation.

Step 2: Calculate the elevation in boiling point (

Δ T_{b}

)○ Expand

For non-electrolytes like ethylene glycol and glucose, the van't Hoff factor ( $i$ ) is 1. The given ebullioscopic constant ( $K_{b}$ ) for water is $0.52 {K kg mol}^{- 1}$ . Using the boiling point elevation formula $Δ T_{b} = i \cdot K_{b} \cdot m$ :

Δ T_{b} = 1 \cdot (0.52 {K kg mol}^{- 1}) \cdot (8 {mol kg}^{- 1}) = 4.16 K

💡 Teacher's Secret Hint

Remember that $i = 1$ for non-dissociating solutes.

Step 3: Determine the boiling point of the solution○ Expand

The normal boiling point of water ( $T_{b}^{0}$ ) is $100^{\circ} C$ , which is $100 + 273.15 = 373.15 K$ . The boiling point of the solution ( $T_{b}$ ) is the sum of the normal boiling point and the elevation in boiling point:

T_{b} = T_{b}^{0} + Δ T_{b} = 373.15 K + 4.16 K = 377.31 K

Rounding to one decimal place, the boiling point of the solution is $377.3 K$ .

💡 Teacher's Secret Hint

Always add the elevation to the normal boiling point of the pure solvent, ensuring consistent units (Kelvin).

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Chemistry Question 54 – JEE-MAIN 2025

Clue 1 of 3: Colligative Properties

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