In titration at equivalence point, number of equivalents of both reactants must be same.

$\therefore {\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2$

$\mathrm{N}$ (normality) of ${\mathrm{FeCl}}_2=\frac{\left(\mathrm{N}\times \mathrm{V}\right) \mathrm{of} {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7}{\mathrm{V} \mathrm{of} {\mathrm{FeCl}}_2}$

$\mathrm{N}$ (normality) of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7=\frac{ \mathrm{Mass} \times 1000}{ \mathrm{Equivalent} \mathrm{mass} \times \mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{prepared}}$

Equivalent mass $=\frac{ \mathrm{Molar} \mathrm{mass} }{\mathrm{n} \mathrm{factor} }$

$=\frac{294}{6}$

$=49 \mathrm{g}/\mathrm{equi}.$

(${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ as oxidising agent is reduced to $2{\mathrm{Cr}}^{3+}$).

Oxidation state of $2$ $\mathrm{Cr}$ atoms change from $+6$ to $+3$.

Total change is$3\times 2=6$, ($\therefore \mathrm{n}$-factor $=6$)

$\therefore \mathrm{N}$ (of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$) $=\frac{1225\times 1000}{49\times 250}=0.1\mathrm{N}$

$\therefore \mathrm{N}$ (of ${\mathrm{FeCl}}_2$) $=\frac{0.1\times 25}{10}$

$=0.25 \mathrm{N}$