Question

One mole of phenol is treated with dilute \(HNO_3\) at \(298 \text{ K}\) 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 ____ \(\times 10^{-1}\) %.
(Given molar mass in \(\text{g mol}^{-1}\) H:1, C:12, N:14, O:16)}

Answer 1

Correct Answer: 175

Step 1: Understanding the Question:
Nitration of phenol with dilute \(HNO_3\) gives ortho-nitrophenol and para-nitrophenol. Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bonding. This is compound (X). We compare the oxygen percentage of (X) with that of phenol.
Step 2: Key Formula or Approach:
\[ \text{% of O} = \frac{\text{Total mass of O in compound}}{\text{Molar mass of compound}} \times 100 \]
Step 3: Detailed Explanation:
1. Phenol (\(C_6H_5OH\)):
Molar mass = \((12 \times 6) + (1 \times 6) + 16 = 72 + 6 + 16 = 94 \text{ g/mol}\).
% of O in phenol = \(\frac{16}{94} \times 100 \approx 17.02%\).
2. Compound X (o-nitrophenol, \(C_6H_5NO_3\)):
Molar mass = \(94 + 14 + (16 \times 2) - 1 = 139 \text{ g/mol}\).
Mass of Oxygen = \(16 \times 3 = 48 \text{ g}\).
% of O in X = \(\frac{48}{139} \times 100 \approx 34.53%\).
3. Increase in percentage:
Increase = \(34.53 - 17.02 = 17.51%\).
Expressed as \(val \times 10^{-1} %\): \(175.1 \times 10^{-1} \approx 175\).
Step 4: Final Answer:
The value is 175.