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`.

Explanation

`(r_(g))/(r_(O_2)) = sqrt((M_(O_2))/(M_((g))))`
` ("LossinPg")/(t_(g))xx (t_(O_2))/("Loss in " P_(O2))= sqrt((32)/(M_(g)))`
`(500)/(85) xx (55)/(500) = sqrt((32)/(M_(g)))`
`M_((g)) =76.43g//mol` .