Percentage Increase of Oxygen in Nitration Product of Phenol

To find the increase in the percentage of oxygen in compound $(\text{X})$ with respect to phenol, we follow these steps:

Step 1: Identify the Steam Volatile Compound (X)

When phenol is treated with dilute $\text{HNO}_3$ at $298\text{ K}$, it undergoes mononitration to yield a mixture of $o$-nitrophenol and $p$-nitrophenol:

$\text{Phenol } (\text{C}_6\text{H}_5\text{OH}) \xrightarrow{\text{dil. }\text{HNO}_3,\ 298\text{ K}} \text{o-nitrophenol} + \text{p-nitrophenol}$

• $o$-Nitrophenol possesses intramolecular hydrogen bonding, which makes it steam volatile.
• $p$-Nitrophenol possesses intermolecular hydrogen bonding, making it non-volatile in steam.

Thus, the steam volatile compound $(\text{X})$ is $o$-nitrophenol ($\text{C}_6\text{H}_5\text{NO}_3$).

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Step 2: Calculate the Percentage of Oxygen in Phenol

• Molecular Formula of Phenol: $\text{C}_6\text{H}_6\text{O}$
• Molar Mass of Phenol: $\text{Molar Mass} = (6 \times 12) + (6 \times 1) + (1 \times 16) = 72 + 6 + 16 = 94\text{ g mol}^{-1}$
• Percentage of Oxygen in Phenol: $\% \text{O}_{\text{phenol}} = \frac{1 \times 16}{94} \times 100 = \frac{1600}{94}\% \approx 17.0213\%$

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Step 3: Calculate the Percentage of Oxygen in Compound (X)

• Molecular Formula of $o$-Nitrophenol: $\text{C}_6\text{H}_5\text{NO}_3$
• Molar Mass of $o$-Nitrophenol: $\text{Molar Mass} = (6 \times 12) + (5 \times 1) + (1 \times 14) + (3 \times 16) = 72 + 5 + 14 + 48 = 139\text{ g mol}^{-1}$
• Percentage of Oxygen in $o$-Nitrophenol: $\% \text{O}_{(\text{X})} = \frac{3 \times 16}{139} \times 100 = \frac{4800}{139}\% \approx 34.5324\%$

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Step 4: Determine the Increase in Percentage of Oxygen

$\text{Increase in } \% \text{ O} = \% \text{O}_{(\text{X})} - \% \text{O}_{\text{phenol}}$

$\text{Increase in } \% \text{ O} = 34.5324\% - 17.0213\% = 17.5111\%$

Expressing this in the form $\text{Value} \times 10^{-1}\%$: $17.5111\% = 175.111 \times 10^{-1}\% \approx 175 \times 10^{-1}\%$

Thus, the required value is 175.