InterviewSolution
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InterviewSolution
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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 isA. `1.44M`B. `0.72M`C. `0.36M`D. `0.8M` |
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Answer» 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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