Nitrobenzene can be prepared from benzene by using a mixture of conc. ${\mathrm{HNO}}_3$ and conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ In the nitrating mixture ${\mathrm{HNO}}_3$  acts as a:

Nitrobenzene can be prepared from benzene by using a mixture of conc. ${\mathrm{HNO}}_3$ and conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ In the nitrating mixture ${\mathrm{HNO}}_3$  acts as a base.

${\mathrm{HNO}}_3$ acts as a base. It reacts with ${\mathrm{H}}_2{\mathrm{SO}}_4$ to form the electrophile ${\mathrm{NO}}_2^{+}$ ions.

${\mathrm{HNO}}_3+2{\mathrm{H}}_2{\mathrm{SO}}_4\rightleftharpoons 2{\left[{\mathrm{HSO}}_4\right]}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{NO}}_2^{+}$

Hence, the option A is correct