Explanation

(c)
Mohr’s salt is $FeSO_{4}.\left( NH_{4} \right)_{2}SO_{4}.6H_{2}O$
Only oxidizable part is $Fe^{2 +}$.
$\lbrack Fe^{2 +} \rightarrow Fe^{3 +} + e^{-}\rbrack \times 6$
$Cr_{2}O_{7}^{2 -} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3 +} + 7H_{2}O$
$6Fe^{2 +} + Cr_{2}O_{7}^{2 -} + 14H^{+} \rightarrow 6Fe^{3 +} + 2Cr^{3 +} + 7H_{2}O$
Millimoles of $Fe^{2 +} = 750 \times 0.6 = 450$
Moles of $Fe^{2 +} = \frac{450}{1000} = 0.450$ mol
6 mol $Fe^{2 +}$=1 mol $Cr_{2}O_{7}^{2 -}$
$\therefore$ 0.450 mol $Fe^{2 +} = \frac{0.450}{6}$
$= 0.075$ mol $Cr_{2}O_{7}^{2 -}$
$= 0.075 \times 294$ g
$= 22.05$ g