The equivalent mass of $K_{2}C{r}_2{O}_7$ when it acts as oxidizing agent in an acidic medium, is equal to _______ fraction of its molecular weight.

$\mathrm{Equivalent} \mathrm{weight} =\frac{\mathrm{molecular} \mathrm{weight}}{\mathrm{Number} \mathrm{of} \mathrm{electrons} \mathrm{lost} \mathrm{or} \mathrm{gained} \mathrm{in} \mathrm{redox} \mathrm{reaction} }$
The molecular weight of = 294 g/mol
Number of electrons gained in acidic medium = 6
Equivalent weight $\frac{294}{6}= 49 \mathrm{g}/\mathrm{eq}$
Hence, the correct answer is 49 g/eq.