In an oxidation-reduction, `MnO_(4)^(-)` ion is converted to `Mn^(2+)`, what is the number of equivalents of `KMnO_(4)` (mol. Wt.=158) present in 250 mL of 0.04 M `KMnO_(4)` solution ?

Options

A. 0.02
B. 0.05
C. 0.04
D. 0.07

Answer

B

Explanation

In redox-reaction :
`8H^(+)+MnO_(4)^(-)+5e^(-) to Mn^(2+)+4H_(2)O`
Change in oxidation state of `MnO_(4)^(-) =(+7)-(+2)=+5`
`therefore N_(KMnO_(4))=M_(KMnO_(4))xx5`
`=0.04xx5=0.20`
Number of equivalents `=(NV)/(1000)=(0.2xx250)/(1000)=0.05`