The pressure in a bulb dropped from `2000` to `1500 mm Hg` in `47 min` when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight `79` in the molar ratio of `1:1` at a total pressure of `4000 mm` of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of `74 min`.

Answer

B

Explanation

The change of pressure of oxygen in `47 min` is `500 mm Hg`. The change of pressure of oxygen after `74 min` is `(500)/(47)xx74=787.2 mm`
In the `1:1` molar ratio mixture of oxygen and another gas, each of them has an equal pressure of `2000mm` because the total pressure is given to be `4000 mm Hg`.
The pressure of oxygen left after `74 min` is
`2000-787.0=1212.8 mm Hg`
`(r_(gas))/(r_(O_(2)))=sqrt((M_(O_(2)))/(M_(gas)))` (Graham's law of diffusion)
`(V_(gas)("diffused"))/(V_(O_(2))("diffused"))xx(t_(O_(2)))/(t_(gas))=sqrt((M_(O_(2)))/(M_(gas)))`
`(P_(gas)("diffused"))/(P_(O_(2))("diffused"))xx(t_(O_(2)))/(t_(gas))=sqrt((M_(O_(2)))/(M_(gas)))`
Both diffuse for the same time, so
`(P_(gas)("diffused"))/(P_(O_(2))(diffused))=sqrt((M_(O_(2)))/(M_(gas)))`
or `(P_(gas)("diffused"))/(787.2)=sqrt((32)/(79))`
`P_(gas)("diffused")=500.8 mm`
The pressure of gas left after `74 min` is
`2000-500.8=1499.2 mm Hg`
Molar ratio of the gas and oxygen left `=(1499.2)/(1212.8)=1.236`