Chemistry · 2026
JEE Main · 5 April 2026, Shift 1 · Q73
One mole of phenol is treated with dilute HNO_3 at 298 K to give a mixture of products. The mixture is separated by steam distillation. The steam…
One mole of phenol is treated with dilute $\displaystyle \mathrm{HNO}_3$ at $\displaystyle 298$ K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound $\displaystyle (\mathrm{X})$ is separated. The increase in percentage of oxygen in $\displaystyle (\mathrm{X})$ with respect to phenol is $\displaystyle \_\_\_\_$ $\displaystyle \times 10^{-1} \%$
(Given molar mass in $\displaystyle \mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16$ )
Official answer
From NTA’s final answer key for this paper.
175
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.