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How many chlorine atoms will be ionised `CI rarr CI^(+) +e^(-1)` by the energy released from the process `CI +e^(-1) rarr CI^(-)` for `6.02 xx 10^(23)` atoms (I.P. for `CI = 1250 kJ mol^(-1)` and `E.A. = 350 kJ "mole"^(-1))` |
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Answer» Correct Answer - `[1.686 xx 10^(23)` atom] `[N_(A) xx E.A = I.P xx n 6.02 xx 10^(23) xx 350 = 1250 xx n]` |
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