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Parveen Prasad Mehrotra
Asked: 2022-11-08T11:18:18+05:30 In:

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

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
A. `1.44M`
B. `0.72M`
C. `0.36M`
D. `0.8M`
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1 Answer

  1. Correct Answer – D
    `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`

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