A mixture containing 1.12 L of `H_(2)` and 1.12 L of `D_(2)` ( deuterium ) at S.T.P. is taken inside a bulb connected to another bulb by a stop-cock with a small opening. The second bulb is fully evacuated, the stop-cock is opened for a certain time and then closed . The first bulb is found to contain -.05 gm of `H_(2)`.Determine the % age composition by weight of the gases in the second bulb.

Explanation

In the first bulb `:`
Initial moles of `H_(2)=1.12 // 22.4 = 1//20`
Initial moles of `D_(2) = 1.12 // 22.4 =1// 20 `
Now after opening of stop-cock,mass of `H_(2)` left in the first bulb `= 0.05`
`rArr` Moles of `H_(2) =0.05 // = 1// 40`
`rArr ` Moles of `H_(2)` effused into second bulb `= 1//20 - 1// 40 = 1//40`
Let n be the number of moles of `D_(2)` effused.
From Graham's law `:`
`( n )/( 1// 40) = sqrt((2 )/( 4))`
`rArr n = ( sqrt(2))/( 80) = ` moles of `D_(2)` is second bulb.
In the second bulb `:`
The mass of `H_(2)` gas `= 1//40 xx 2=0.05 ` gm
The mass of `D_(2)` gas `= sqrt( 2) //80 xx 4 = 0.07 ` gm
`rArr` Total mass `= 0.05 + 0.07 = 0.12 gm`
`rArr ` % of `H_(2)=0.05 // 1.12 xx 100 =41.67 %`
`rArr %` of `D_(2) = 0.07 // 1.12 xx100 = 58.13 %`