A glass bulb contains 2.24 L of H_(2) and 1.12 L of D_(2) at S.T.P. It is connected to a fully evacuated bulb by a stocock with a small opening. The stopcock is opened for some time and then closed. The first bulb now contains 0.10 g of D_(2).Calculate the percentage composition by weight of the gases in the second bulb.

A glass bulb contains 2.24 L of H_(2) and 1.12 L of D_(2) at S.T.P. It

1.	A glass bulb contains 2.24 L of H_(2) and 1.12 L of D_(2) at S.T.P. It is connected to a fully evacuated bulb by a stocock with a small opening. The stopcock is opened for some time and then closed. The first bulb now contains 0.10 g of D_(2).Calculate the percentage composition by weight of the gases in the second bulb.
Answer» SOLUTION :Weight of 2.24 L of `H_(2)` at S.T.P.=0.2 g 0.1 mol(Mol. Mass of `H_(2)`=2) Weightof 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%`