The reaction of Miohr's salt and $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is as follow:
$6 \mathrm{Fe}^{2+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14\left[\mathrm{H}^{+}\right] \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$ $n$-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 $0.08 \mathrm{M} \mathrm{Mohr's} \mathrm{salt} \mathrm{}$
$$
=0.08 \times \frac{25}{1000}=0.002
$$
$\therefore$ Moles in $500 \mathrm{cc}$ of Mohr's salt
$$
=0.002 \times \frac{500}{25}=0.04
$$
Hence, mass of Mohr's salt in $500 \mathrm{cc}$
$$
=0.04 \times 392=15.68 \mathrm{~g}
$$