The reaction of Miohr's salt and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is as follow:
$6{\mathrm{Fe}}^{2+}+{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14\left[{\mathrm{H}}^{+}\right]⟶6{\mathrm{Fe}}^{3+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$ $n$-factor of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ is 6 and $n$-factor of ${\mathrm{Fe}}^{2+}$ is 1 .
As $n$-factor for $\mathrm{Fe}$ is 1 so,
Molarity = Normality Concentration of Mohr's salt $=0.08\mathrm{M}$
Moles in $25\mathrm{mL}$ of $0.08\mathrm{M}{\mathrm{Mohr}}^{\prime }\mathrm{s}\mathrm{salt}$
$$=0.08\times \frac{25}{1000}=0.002$$
$\therefore$ Moles in $500\mathrm{cc}$ of Mohr's salt
$$=0.002\times \frac{500}{25}=0.04$$
Hence, mass of Mohr's salt in $500\mathrm{cc}$
$$=0.04\times 392=15.68\mathrm{g}$$