How many milliliters fo a `0.05 M KMnO_4` solution are required to oxidize `2.0 g FeSO_4` in a dilute acid solution ?A. ` 32. 56 mL`B. `62.53 mL`C. ` 25.36mL`D. ` 52.63 mL`

How many milliliters fo a `0.05 M KMnO_4` solution are required to oxi

1.	How many milliliters fo a `0.05 M KMnO_4` solution are required to oxidize `2.0 g FeSO_4` in a dilute acid solution ?A. ` 32. 56 mL`B. `62.53 mL`C. ` 25.36mL`D. ` 52.63 mL`
Answer» 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`.

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How many milliliters fo a `0.05 M KMnO_4` solution are required to oxidize `2.0 g FeSO_4` in a dilute acid solution ?A. ` 32. 56 mL`B. `62.53 mL`C. ` 25.36mL`D. ` 52.63 mL`

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