A glass bulb contains `2.24L of H_(2)` and `1.12L` of `D_(2)` at `STP`. It is connected a fully evacuated bulb by a stop-cock with a small opening. The stop-cock is opened for sometime and then closed. The first bulb now contains `0.10 g` of `H_(2)`. The percentage of `H_(2)` in the mixture is

Explanation

Weight of 2.24 L of `H_(2)` at S.T.P.=0.2 g
0.1 mol (Mol. Mass of `H_(2)`=2)
Weight of 1.12 L of `D_(2)` at S.T.P.=0.2 g
=0.05 mol ( Mol. Mass of `D_(2)`=4)
As number of moles of the two gases are different but V and T are same, therefore, their partial pressures will be different, i.e., in the ratio of their number of moles. Thus,
`(P_(H_(2)))/(P_(D_(2)))=(n_(H_(2)))/(n_(D_(2)))=(0.1)/(0.05)=2`
Now, `D_(2)` present in the first bulb=0.1 g (Given)
`:. D^(2)` diffused into the second bulb `=0.2-0.1=0.1" g"=0.56" g L"` at S.T.P.
Now, `" " (r_(H_(2)))/(r_(D_(2)))=(P_(H_(2)))/(P_(D_(2)))xxsqrt((M_(D_(2)))/(M_(D_(2))))" "or " "(v_(H_(2)))/(t)xx(t)/(v_(D_(2)))=(P_(H_(2)))/(P_(D_(2)))xxsqrt((M_(D_(2)))/(M_(H_(2))))`
`(v_(H_(2)))/(t)xx(t)/(0.56" L")=2xxsqrt((4)/(2))" or " v_(H_(2))=1.584" L "=014" g"` of `H_(2)`
`(`:'` 22.4" L "H_(2)=2" g "H_(2))`
`:.` Weight of the gases in 2nd bulb `=0.10 g (D_(2))+0.14 g(H_(2))=0.24" g"`
Hence, in the 2nd bulb,
% of `D_(2)` by weight `=(0.10)/(0.24)xx100=14.67%`
% of `H_(2)` by weight `=100-41.67=58.33%`