${\mathrm{P}}_4+3\mathrm{NaOH}+3{\mathrm{H}}_2\mathrm{O}\to {\mathrm{PH}}_3+3{\mathrm{NaH}}_2{\mathrm{PO}}_2$  The equivalent mass of ${\mathrm{P}}_4$ in the above reaction is (M = Molecular weight of ${\mathrm{P}}_4$) 

$12{\mathrm{e}}^{-1}+{\mathrm{P}}_4\to 4{\mathrm{P}}^{-3}$

${\mathrm{P}}_4\to 4{\mathrm{P}}^{-3}+4{\mathrm{e}}^{-1}$

$\because 4\mathrm{mole}{\mathrm{P}}_4\Rightarrow 12\mathrm{mole}{\mathrm{e}}^{-1}$ gain or loss 

$\therefore$1 mole ${\mathrm{P}}_4\Rightarrow \frac{12}{4}=3\mathrm{mole}{\mathrm{e}}^{-1}$

Equivalent weight $=\frac{\mathrm{M}}{3}$