When 2g of a gas A is introduced into an evacuated flask kept at `25^@C`, the pressure is found to be one atmosphere. If 3 g of another gas B are then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of molecular weights `M_A : M_B`.

Options

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

Answer

A

Explanation

Moles `prop` Pressure
`(2)/(Mw_(A)) prop 1 "atm"`
Pressure of `B = 1.5 - 1 = 0.5 "atm"`
`(3)/(Mw_(B)) prop 0.5 "atm"`
`(3)/(Mw_(B)) xx (Mw_(A))/(2) = (0.5)/(1)`
`(MW_(A))/(Mw_(B)) = 0.5 xx (2)/(3) = (1)/(3)`
`(Mw_(A) : Mw_(B) = 1:3`