What is the equivalent weight of ${\mathrm{HNO}}_3$ in the given reaction ?

$4\mathrm{Zn} + 10{\mathrm{HNO}}_3 \to 4\mathrm{Zn}{\left({\mathrm{NO}}_3\right)}_2 + {\mathrm{NH}}_4{\mathrm{NO}}_3 + 3{\mathrm{H}}_2\mathrm{O}$

(3)

${\mathrm{HNO}}_3$$\to$${\mathrm{NH}}_4^{+}$

${\mathrm{N}}^{5+} + 8{\mathrm{e}}^{-} \to {\mathrm{N}}^{3-}$

n-factor as reductant=8

9 moles of HNO₃ is acting as acid, basicity is 1. 

As an acid, equivalent weight = M

Total equivalent weight = M (as acid) +$\frac{\mathrm{M}}{8}$( as Reducing agent)=$\frac{9\mathrm{M}}{8}$=$\frac{9\times 63}{8}$