Chemistry · 2026
JEE Main · 21 January 2026, Shift 2 · Q71
A substance ' X ' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass =300 g mol^-1 ) led to an elevation of the boiling point by 0.5 K. The…
A substance ' $\displaystyle X$ ' ($\displaystyle 1.5$ g) dissolved in $\displaystyle 150$ g of a solvent 'Y' (molar mass $\displaystyle =300 \mathrm{~g} \mathrm{~mol}^{-1}$ ) led to an elevation of the boiling point by $\displaystyle 0.5$ K. The relative lowering in the vapour pressure of the solvent 'Y' is $\displaystyle \_\_\_\_$ $\displaystyle \times 10^{-2}$. (nearest integer)
[Given : $\displaystyle \mathrm{K}_{\mathrm{b}}$ of the solvent $\displaystyle =5.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ]
Assume the solution to be dilute and no association or dissociation of $\displaystyle X$ takes place in solution.
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From NTA’s final answer key for this paper.
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.