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. `1.44M`
B. `0.72M`
C. `0.36M`
D. `0.8M`

Answer

D

Explanation

`CN^(-) +5H_(2)O rarr CO_(2) +NO_(3)^(-) +10H^(+) +10e`
`MnO_(4)^(-) +8H^(+) +5e^(-) rarr Mn^(2+) +4H_(2)O`
`ulbar(CN^(-)+2MnO_(4)^(-)+6H^(+)rarr CO_(2)+NO_(3)^(-)+2Mn^(2+) +3H_(2)O)`
`ca(OH)_(2) +CO_(2) rarr CaCO_(3)+H_(2)O`
`100g` of `CaCO_(3) -= 44g` of `CO_(2)`
`1g` of `CaCO_(3) -= 0.44g` of `CO_(2) = 0.01` moles of `CO_(2)`
1 mole of `CO_(2) -= 2` moles `MnO_(4)^(-)`
`0.01` moles of `CO_(2) -= 0.02` moles of `MnO_(4)^(-)`
Molarity of `KmnO_(4) = (0.02 xx 1000)/(25) = 0.8M`