The equivalent weight of ${\mathrm{MgCl}}_2 \mathrm{is}$

Equivalent weight of ${\mathrm{MgCl}}_2=\frac{\mathrm{Formula} \mathrm{weight} \mathrm{of} {\mathrm{MgCl}}_2}{\mathrm{Charge} \mathrm{of} \mathrm{cation} \mathrm{or} \mathrm{anion}}$

In ${\mathrm{MgCl}}_2$ ,there are one ${\mathrm{Mg}}^{+2}$and two -${\mathrm{Cl}}^{-}$ions are present.

Then, the charge on cation ${\mathrm{Mg}}^{+2} \mathrm{is} 2$

The formula weight of ${\mathrm{MgCl}}_2 \mathrm{is} 95 \mathrm{g}.$

Therefore, Equivalent weight of ${\mathrm{MgCl}}_2=\frac{95}{2}=47.5$