In the reaction:

$\mathrm{Zn}+{\mathrm{HNO}}_3\to \mathrm{Zn}{\left({\mathrm{NO}}_3\right)}_2+\mathrm{NO}+{\mathrm{H}}_2\mathrm{O}$, the equivalent weight of HNO₃ is (Molar mass of HNO₃ = M)

3 electrons are gain
$$\begin{gathered} \mathrm{Eq}=\frac{\mathrm{molar}\mathrm{mass}}{\mathrm{nF}}\mathrm{nF}=3 \\ \begin{array}{l}=\frac{4\mathrm{M}}{3} \underset{ }{8\mathrm{molecule}\mathrm{of}{\mathrm{HNO}}_3}\to 2\mathrm{molecule}\mathrm{of}\mathrm{NO}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\mathrm{molecule}\mathrm{of}{\mathrm{HNO}}_3\to \mathrm{react}\mathrm{with}\frac{2}{8}\times 3=\frac{3}{4}=\mathrm{nF}\end{array} \end{gathered}$$