JEE Advanced 2024 — Paper 2 — CHEM-III solution

JEE Advanced2024Paper 2CHEM-III

Vessel-1 contains $w_{2}$ g of a non-volatile solute $X$ dissolved in $w_{1}$ g of water. Vessel-2 contains $w_{2}$ g of another non-volatile solute $Y$ dissolved in $w_{1}$ g of water. Both the vessels are at the same temperature and pressure. The molar mass of $X$ is $80%$ of that of $Y$ . The van't Hoff factor for $X$ is $1.2$ times of that of $Y$ for their respective concentrations.

The elevation of boiling point for solution in Vessel-1 is _____ $%$ of the solution in Vessel-2.

Your answer

Solution

For a non-volatile solute, the boiling point elevation is $Δ T_{b} = i \cdot K_{b} \cdot m$ , where the molality $m = \frac{w _{2}}{M \cdot w _{1}} \times 1000$ depends on the molar mass $M$ of the solute.

For Vessel-1 (solute $X$ ): $Δ T_{b, 1} = i_{X} \cdot K_{b} \cdot \frac{w _{2}}{M _{X} \cdot w _{1}} \times 1000.$

For Vessel-2 (solute $Y$ ): $Δ T_{b, 2} = i_{Y} \cdot K_{b} \cdot \frac{w _{2}}{M _{Y} \cdot w _{1}} \times 1000.$

Taking the ratio (everything except $i$ and $M$ cancels):

$\frac{Δ T _{b, 1}}{Δ T _{b, 2}} = \frac{i _{X}}{i _{Y}} \cdot \frac{M _{Y}}{M _{X}} .$

Given $i_{X} = 1.2 i_{Y}$ and $M_{X} = 0.8 M_{Y}$ :

$\frac{Δ T _{b, 1}}{Δ T _{b, 2}} = 1.2 \times \frac{1}{0.8} = 1.2 \times 1.25 = 1.5.$

$∴ Δ T_{b, 1}$ is $150%$ of $Δ T_{b, 2} .$

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