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ChemistryInteger+4 / -1Question 73 of 75

JEE Main 2026 · 5 Apr · Shift 1, Chemistry question 73

One mole of phenol is treated with dilute HNOX3\ce{HNO3} at 298 K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound (X) is separated. The increase in percentage of oxygen in (X) with respect to phenol is ________ ×10−1\times 10^{-1} %

(Given molar mass in g mol−1^{-1} H:1, C:12, N:14, O:16)

Answer and worked solution

Official answer: 175

X. Dilute HNOX3\ce{HNO3} gives o- and p-nitrophenol. The ortho isomer has an intramolecular hydrogen bond, so it is the steam-volatile one: X is o-nitrophenol, CX6HX5NOX3\ce{C6H5NO3} (M = 139).

Oxygen content.

  • X: 48139×100=34.53%\tfrac{48}{139}\times100=34.53\%.
  • Phenol, CX6HX6O\ce{C6H6O} (M = 94): 1694×100=17.02%\tfrac{16}{94}\times100=17.02\%.

Increase: 17.51%≈175×10−1%17.51\%\approx175\times10^{-1}\%.

Answer: 175

Topics:PhenolsEmpirical and molecular formula
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