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. `1:1`
B. `1:2`
C. `2:3`
D. `1:4`

Answer

C

Explanation

After the introduction of `3 g` of `B` the pressure became half its number of mole of `B`
`PV=nRT`
`n=(w)/(M)implies (w_(A))/(M_(A))=(w_(B))/(M_(B))`
or `(2)/(M_(A))=(3)/(M_(B)) or (M_(A))/(M_(B))=(2)/(3)`