A glass bulb contains `2.24 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

Options

A. `41.6%`
B. `58.4%`
C. `46.2%`
D. `50%`

Answer

A

Explanation

`1 "mole"` of `H_(2)(2g)` occupies `22.4L`
`2.24 L` of `H_(2)` at STP`=(2)/(22.4)xx2.24= 0.2g`
`1.12L` of `D_(2) (4g)` at STP`=(4)/(22.4)xx1.12= 02g`
Mass of `H_(2)` diffused in given time
`= 0.2 g- 0.1 g= 0.1 g`
`(m_(H_(2)))/(m_(D_(2)))=sqrt((M_(H_(2)))/(M_(D_(2))))`
`m_(D_(2))=sqrt((M_(D_(2)))/(M_(H_(2))))xxm_(H_(2))=sqrt((4)/(2))xx0.1g= 0.14 g`
`% of H_(2)` in this bulb `=(0.1)/(0.1+0.14)xx100=41.6%`