InterviewSolution
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InterviewSolution
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One mole of `A` present in a closed vessel undergoes decay as: `._(Z)^(A) rarr ._(Z-4)^(m-8)B + 2 ._(2)^(4)He`. The volume of `He` collected at `NTP` after 20 day (`t_(1//2)` for `A = 10` day) is:A. `11.2` litreB. `22.4` litreC. `33.6` litreD. `67.2` litre |
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Answer» Correct Answer - `(c)` Amount left `= (N_(0))/(2^(2)), (t_(1//2) xx n = T)` `:.` Amount decayed `= N_(0) - (N_(0))/(4) = (3N_(0))/(4)` Also 1 mole of `A` gives two moles of `He` thus moles of `He` formed `= (3)/(4)xx 2 = (3)/(2)` mole `= (3)/(2) xx22.4` litre `= 33.6` litre. |
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