Correct Answer – D
Normality `=n_(“factor”)xx`Molarity
`overset(+7)(M)nO_(4)^(-) rarroverset(+2)(M)n^(2+), n_(“factor”)=5`
`Fe^(2+) rarr Fe^(3+), n_(“factor”)=1`
Normality of `KMnO_(4)=(5)(0.05)`
`=0.25 N`
Volume of `KMnO_(4)=V` milliliters
Thus, milliequivalents of `KMnO_(4)=NxxV`
`=0.25 V`
Equivalents of `FeSO_(4)=(“Mass”_(FeSO_(4)))/(“Gram equivalent mass”_(FeSO_(4)))`
`[“Note that eq. wt. of “FeSO_(4)=(“Formula weight”)/(“Change in O.N.”)=152/1]`
Milliequivalent of `FeSO_(4)=2/152xx1000`
According to the law of equivalence,
`”Milliequivalents”_(KMnO_(4))=”Miliequivalents”_(FeSO_(4))`
`0.25 V= 2/(152) xx 1000`
` V= (2xx 1000)/(152 xx 0. 25) `
` = 52. 63 mL`.