InterviewSolution
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Auto-ionisation of liquid `NH_(3)` is `2NH_(3) hArr NH_(4)^(o+) +NH_(2)^(Theta)` with `K_(NH_(3)) = [NH_(4)^(o+)] [NH_(2)^(Theta)] = 10^(-30) at -50^(@)C` Number fo amide ions `(NH_(2)^(Theta))`, present per `mm^(3)` of pure liquied `NH_(3)` isA. `602`B. `301`C. `200`D. `100` |
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Answer» Correct Answer - A `2NH_(3) hArr NH_(4)^(o+) +NH_(2)^(Theta)` `[NH_(4)^(o+)] = [NH_(2)^(Theta)]= sqrt(K_(NH_(3))) = 10^(-15)M` `1dm^(3)` of solution contains `10^(-15)` moles `= 6.02 xx 10^(23) xx 10^(15)` molecules of `NH_(2)^(Theta)`. `1mm^(3)` of solution contains `6.02 xx 10^(23) xx 10^(-15) xx 10^(-6) = 602 NH_(2)^(Theta)` ions. |
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