One mole of phenol is treated with dilute HNO3 at 298 K to give a mixture of products. The mixture is separated by steam distillation. The seam volatile compound ( X ) is separated. The increase in percentage of oxygen in (X) with respect to phenol is ____ 10−1 % (Given molar mass in gmol−1H:1,C:12,N:14, O:16)
Answer: 175
Solution
Nitration of phenol with dilute nitric acid at 298K gives a mixture of ortho- and para-nitrophenol, and of these the ortho isomer is steam-volatile, since its intramolecular hydrogen bond prevents the intermolecular association that would hold it back, so X is o-nitrophenol. In phenol, C6H6O of molar mass 94, the mass fraction of oxygen is 9416×100=17.02%. In o-nitrophenol, C6H5NO3 of molar mass 139, it is 13948×100=34.53%. The rise is 34.53−17.02=17.51%, equal to 175×10−1%, so the nitrogen percentage is 175.