When `2 g` of a gas `A` is introduced into an evacuated flask kept at `25^(@)C`, the pressure is found to be `1 atm`. If `3 g` of another gas `B` is then heated in the same flask, the total pressure becomes `1.5 atm`. Assuming ideal gas behaviour, calculate the ratio of the molecular weights `M_(A)` and `M_(B)`.

Options

A. `1 :3`
B. `1 :4`
C. `4 : 1`
D. `3 :1`

Answer

A

Explanation

Partial pressure of gas A in the mixture of = 1 atm
Partial pressure of gas B in the mixture = 0.5 atm
we know that
`(P_(A))/(P_(B)) = (X_(A))/(X_(B)) = (n_(A))/(n_(B))`
`therefore " " (1)/(0.5) = (2 xx M_(B))/(M_(A) xx 3) = (M_(A))/(M_(B)) = (0.5 xx 2)/(3) = (1)/(3)`
`therefore " " M_(A) : M_(B) = 1 : 3`