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`.

Answer

`(1 : 3)`

Explanation

From the given information, it can be easily deduced that in the final mixture,
Partial pressure of A = 1.0 atm, Partial pressure of B = 0.5 atm
Also `n_(A) = (p_(A)V)/(RT) = (V)/(RT), n_(B) = (p_(B)V)/(RT) = (0.5V)/(RT) rArr (n_(B))/(n_(A)) = (1)/(2) (w_(B))/(M_(B)) xx (M_(A))/(w_(A)) = (3)/(2) xx ((M_(A))/(M_(B))) rArr M_(A) : M_(B) = 1 : 3`