One mole of a present in a closed vessel undergoes decay as `""_(z)""^(m)Ato""_(Z-4)""^(m-8)B+2""_(2)""^(4)He`. The volume of He collected at NTP a fter 20 days is (`t""_(1//2)=10`days) a)11.2 litre b)22.4 litre c)33.6 litre d)67.2 litre

Explanation

We know that,
`N = N_(0) ((1)/(2))^(n)` N = remaining mole of A
`N = 1 ((1)/(2))^(2) = (1)/(4)`
Number of decayed moles `= 1 - (1)/(4) = (3)/(4)`
Number of moles of helium formed
`= 2 xx` Number of decayed moles of `A = 2 xx (3)/(4) = (3)/(2)`
Volume of helium at STP `= (3)/(2) xx 22.4 = 33.6` litre