Explanation

(d)
Equivalent weight of
$K_{2}{Cr}_{2}O_{7}\ \ = \ \frac{molecular\ weighty\ of\ K_{2}{Cr}_{2}O_{7}\ }{oxidation\ number\ of\ Cr}$
Oxidation number of Cr in $K_{2}{Cr}_{2}O_{7}$
2[+1]+2(x)+7(-2)= 0
2+2x-14 = 0
2x=12
x=6
Equivalent weight =$\frac{294.19}{6} = 49.08$
$\frac{weight\ of\ K_{2}{Cr}_{2}O_{7}\ \ }{equivalent\ wt.(E)} = N \times V(L)\ \ \ \$
$w = 0.1 \times \ 1\ \times 49.03$ = 4.903 g