If in a reaction ${\mathrm{HNO}}_3$ is reduced to NO, the mass of ${\mathrm{HNO}}_3$absorbing one mole of electrons would be 

We have  ${\mathrm{HNO}}_3⟶+{\mathrm{N}}^{+2}\mathrm{O}$

Change m oxidation number= -3. Equivalent mass of ${\mathrm{HNO}}_3=\frac{62\mathrm{g}{\mathrm{mol}}^{-1}}{3\mathrm{cq}{\mathrm{mol}}^{-1}}=21{\mathrm{geq}}^{-1}$