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One mole of A present in a closed vessel...

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` litre

B

`22.4` litre

C

`33.6` litre

D

`67.2` litre

Text Solution

Verified by Experts

The correct Answer is:
`(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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