Chemistry · 2026
JEE Main · 4 April 2026, Shift 1 · Q75
A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750…
A non-volatile, non-electrolyte solid solute when dissolved in $\displaystyle 40$ g of a solvent, the vapour pressure of the solvent decreased from $\displaystyle 760$ mm Hg to $\displaystyle 750$ mm Hg . If the same solution boils at $\displaystyle 320$ K , then the number of moles of the solvent present in the solution is $\displaystyle \_\_\_\_$. (Nearest integer)
[Given: boiling point of the pure solvent $\displaystyle =319.5 \mathrm{~K}$, $\displaystyle \mathrm{K}_{\mathrm{b}}$ of the solvent $\displaystyle \left.=0.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$
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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.