A mixture containing `1.12` litre `D_(2)` and 2.24 liter of `H_(2)` at `NTP` is taken inside a bulb connected to another bulb through a stop cock with a small opening The second bulb is fully evacuated The stop cock is opened for a crtian time and then closed The first bulb is now found to contain `0.10g` of `D_(2)` Determine the % by weight of the gases in second bulb
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Explanation

At `STP` Before diffusion
`{:(D_(2)=1.12lit.atSTP=0.2g=0.05"mole"),(H_(2)=2.24lit.atSTP=0.2g=0.1"mole"):}}"inIbulb"`
When these moles are placed in the bulb, the partial pressure of gas will be different because `V` and `T` are constant Also `P prop n`
Thus `P_(D_(2))/P_(H_(2))=(0.05)/(0.10)=(1)/(2)`
After diffusion `D_(2)` left in I bulb `=0.1g`
or `D_(2)` diffuses from I into II (bulb) `=0.2 -0.1 =0.1`
Now for diffusion of `D_(2)` and `H_(2)`
`(r_(D_(2)))/(r_(H_(2)))=sqrt(((M_(H_(2)))/(M_(D_(2)))))xx(P_(D_(2)))/(P_(H_(2)))`
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``:'w_(D_(2))/(t_(D_(2)))xx(t_(H_(2)))/(w_(H_(2)))=sqrt(((M_(D_(2)))/(M_(H_(2)))))xx(P_(D_(2)))/(P_(H_(2)))`
`:. (0.1)/(t) xx (t)/(w_(H_(2)))=sqrt(((4)/(2)))xx(1)/(2)`
`w_(D_(2)) =0.14g`
`:.` Wt. of gases in II bulb =wt. of `D_(2) +`wt.of `H_(2)`
`=0.10 g + 0.14g = 0.24 g`
`:. % D_(2) hy wt =(0.10)/(0.24)xx 100 =41.66%`
`%H_(2)` in bulb II `=58.38%` .