Naturally occurring hdrogen gas has abundance of `._(1)^(1)H_(1)^(1)` (mol.wt.2) and `._(1)^(1)H_(1)^(2)H` (mol. Wt. 3). The % of `.^(1)H^(1) H = 90.0%`. In now much operations during diffusion of natural gas will be entriched to `99.8% .^(1)H^(1)H`?

Explanation

`H_(2)` present has `._(1)^(1)H_(1)^(1)H` and `._(1)^(1)H_(1)^(2)H` in the ratio `99.98%` and `0.2` i.e `n_(2_(H)):n_(3_(H))` This has to deffused to have the outcoming sample enriched to `99.8% .^(1)H^(1)H`
`n_(2_H)=90, n_(3_(H))=10 :. (n_(A))/(n_(B)) =(90)/(10)`
`n'_(2_(H))=99.8,n'_(3_(H)) =0.2 :. (n'_A)/(n'_B) = (99.8)/(0.2)`
`:.` Separation factor `= (99.8 xx 10)/(0.2 xx 90) =55.44`
`:. f=(r_(2_(H)))/(r_(3_(H))) =sqrt((M_(3_(H)))/(M_(2_(H)))) =sqrt((3)/(2))`
Now `(f')^(X) = [(n'_A//n'_B)/(n_(A)//n_(B))]`
`:. [sqrt((3)/(2))]^(X)=55.44`
`:. (X)/(2) log .(3)/(2) =55.44`
`:. X =19.8 ~ 20` steps .