How many mL. of aqueous solution of `KMnO_(4)` containing `158g/L` must be used to complete the conversation of `75.0g` of `KI` to iodine by the reaction
`KMnO_(4)+KI+H_(2)SO_(4)rarrK_(2)SO_(4)+MnSO_(4)+I_(2)+6H_(2)O`

Explanation

`underset(315.8 g)(2KMnO_(4)) + 8H_(2)SO_(4) + underset(1665 g)(10 KI) to 6K_(2)SO_(4) + 2MnSO_(4) + 5I_(2) + 8H_(2)O`
1665 g of KI require for conversion to `I_(2) = 315.8`g
`KMnO_(4)`
75 g of Kl will require for conversion to `I_(2)`
`=(315.8 xx 75)/1665 = 14.22`g
158 g of `KMnO_(4)` is present in =1000 mL
14.23 g of `KMnO_(4)` will be present in
`=1000/158 xx 14.22 = 90.05` mL