Certain " mol of "HCn is oxidised completely by 25 " mL of " `KMnO_(4)`. The products are `CO_(2)` and `NO_(3)^(ɵ)` ion. When all `CO_(2)` is passed through lime water , 1 g of `CaCO_(3)` is obtained the molarity of the `KMnO_(4)` used is

Options

A. 1.44 M
B. 0.72 M
C. 036 M
D. none of these

Answer

D

Explanation

`(1 g) -= 0.01 mol`
mol of `CO_(2) = 0.01` mol
`CN^(ɵ)+5H_(2)OtoCO_(2)+NO_(3)^(ɵ)+10H^(o+)+10e^(-)`
`underline(MnO_(4)^(ɵ)+8H^(o+)+5e^(-)toMn^(2+)+4H_(2)O)`
`underline(CN^(ɵ)+2MnO_(4)^(ɵ)+6H^(o+)toCO_(2)+NO_(3)^(ɵ)+2Mn^(2+)+3H_(2)O)` ltBrgt `1 mol CO_(2)-=2 mol MnO_(4)^(ɵ)`
`0.01 mol CO_(2)-=0.02 mol MnO_(4)^(ɵ)`
`M_(KMnO_(4))=(0.02)/((25)/(1000))=0.8M`