Explanation

(d)
Equivalent weight
$= \frac{Molecular\ mass}{Change\ in\ oxidation\ number\ per\ atom}$
or $Cr_{2}O_{7}^{2 -} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3 +} + 7H_{2}O$
Equivalent weight of $K_{2}Cr_{2}O_{7}$
$= \frac{molecular\ weight\ of\ K_{2}Cr_{2}O_{7}}{2 \times change\ in\ oxidation\ number}$
$= \frac{M}{2 \times 3}$ $\lbrack\because$ Two Cr atoms are involved.$\rbrack$
$\therefore$ Equivalent weight of $K_{2}Cr_{2}O = \frac{M}{6}$