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)

Step 1: Identify Compound (X)

• When phenol is treated with dilute $$HNO_3$$ at $$298 K$$, it undergoes nitration to give a mixture of o-nitrophenol and p-nitrophenol.
• On steam distillation, o-nitrophenol is separated because it is steam volatile due to intramolecular hydrogen bonding. 
• Therefore, compound (X) is o-nitrophenol. 

Step 2: Oxygen Percentage in Phenol ($$C_{6}H_{5}OH$$)

Molar Mass: $$6(12)+5(1)+1(16)+1=94$$

Percentage of Oxygen :

$$\frac{16}{94}\times\ 100=17.02\ \%\ $$

Step 3: Oxygen Percentage in o-nitrophenol ($$C_{6}H_{5}NO_3$$)

Molar Mass: $$6(12)+5(1)+14+3(16)=139$$

Percentage of Oxygen:$$\frac{48}{139}\times\ 100=34.53\ \%\ $$

Step 4: Increase in Oxygen Percentage

$$34.53\ -17.02=17.51$$

so,

$$17.51\ \%\ =175.1\times\ 10^{-1}\ \%\ $$

Therefore,

The increase in percentage of oxygen in (X) with respect to phenol is 175 $$\times 10^{-1}$$ %.