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 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 g mol$^{-1}$ H:$1$, C:$12$, N:$14$, O:$16$)

When phenol is treated with dilute $\text{HNO}_3$ at $298\text{ K}$, it undergoes electrophilic aromatic substitution to give a mixture of ortho-nitrophenol and para-nitrophenol.

The mixture is separated by steam distillation. Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bonding, whereas para-nitrophenol is not steam volatile due to intermolecular hydrogen bonding. Thus, the steam volatile compound (X) is ortho-nitrophenol.

Molecular formula of phenol: $\text{C}_6\text{H}_6\text{O}$
Molar mass of phenol $= (6 \times 12) + (6 \times 1) + 16 = 94\text{ g mol}^{-1}$
Percentage of oxygen in phenol $= \dfrac{16}{94} \times 100 \approx 17.02\%$

Molecular formula of ortho-nitrophenol (X): $\text{C}_6\text{H}_5\text{NO}_3$
Molar mass of ortho-nitrophenol $= (6 \times 12) + (5 \times 1) + 14 + (3 \times 16) = 139\text{ g mol}^{-1}$
Percentage of oxygen in ortho-nitrophenol $= \dfrac{48}{139} \times 100 \approx 34.53\%$

Increase in percentage of oxygen $= 34.53\% - 17.02\% = 17.51\%$

We need to express this as $y \times 10^{-1} \%$.
$17.51\% = 175.1 \times 10^{-1} \%$

Rounding to the nearest integer, we get $175$.

Answer: $175$