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

Calculation of amount of `H_(2)` in the bulb
`"22.4 of "H_(2)" at S.T.P. weight "=2g`
`"2.24 L of "H_(2)" at S.T.P. weigh "=(2xx2.24)/(22.4)=0.2g`
Similarly,
Amount of `D_(2)` in bulb `=(1.12xx4)/(22.4)=0.2g`
Amount of `D_(2)` left after opening the stop cock `=1.10g`
`therefore" Amount of "D_(2)" diffused"=0.2-1.10=0.1g`
`"Volume of "O_(2)" diffused "=(0.1xx22.4)/(4)=0.56L`
Let the volume of `H_(2)` diffused from first bulb = x L
According the Graham.s law,
`"rate, r"=(V)/(t)`
`r_(H_(2))=(x)/(t) and r_(D_(2))=(0.56)/(t)`
`therefore" "(r_(H_(2)))/(r_(D_(2)))=((x)/(t))/((0.56)/(t))`
`"Now, "((x)/(t))/((0.56)/(t))=sqrt((M_(D_(2)))/(M_(H_(2))))=sqrt((4)/(2))`
`therefore" "x=0.56xxsqrt2`
`=0.56xx1.414=0.79L`
`"Amount of "D_(2)" in the second bulb = 0.1 g"`
Amount of `H_(2)` in the second bulb `=(0.79xx2)/(22.4)`
`=0.0707g`
Total amount of gases in the second bulb
`=0.1+0.0707`
`%" of "H_(2)" gas"=(0.0707xx100)/(0.1707)`
`=41.2%`
`%" of "D_(2)" gas "=(0.1xx100)/(0.1707)=58.8%`.