When 2.0 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 them 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).

When 2.0 g of a gas A is introduced into an evacuated flask kept at 25

1.	When 2.0 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 them 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» Solution :Total pressure = 1.5 ATM. Moles of A `= (2)/(M_(A))` Moles of B `= (3)/(M_(B))` Total moles `= (2)/(M_(A)) + (3)/(M_(B))` Partial pressure of A `= (2//M_(A))/(2//M_(A) + 3//M_(B)) XX 1.5` and partial pressure of B `= (3//M_(B))/(2//M_(A) + 3//M_(B)) xx 1.5` As GIVEN, partial pressure of A = 1 atm. Partial pressure of B = 1.5 - 1 = 0.5 atm. THUS, `((2//M_(A))/(2//M_(A)+3//M_(B)) xx 1.5)/((3//M_(B))/(2//M_(A) + 3//M_(B)) xx 1.5) = (1)/(0.5) = 2` or `(2)/(3) xx (M_(B))/(M_(A)) = 2` or `(M_(A))/(M_(B)) = (1)/(3)`