Equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ in the following reaction is 

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-} + {\mathrm{Fe}}^{2+} \to {\mathrm{Fe}}^{3+}+ {\mathrm{Cr}}^{3+}$

(M = molar mass of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$)

Balance Cr oxidation number of Cr-atom on each side

Thus, equivalent weight of

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ = $\frac{\mathrm{molar} \mathrm{mass}}{\mathrm{change} \mathrm{in} \mathrm{ox}.\mathrm{state}} = \frac{\mathrm{M}}{6}$