When 2 grams of a gas A is introduced in an evacuated flask kept at 25 C; the pressure was found to be 1 atm if 3 g of another gas B is then added to the same flask; the pressure becomes 1.5 atm.Assuming idel behaviour ; find ratio of molecular weights(MA:MB)

When 2 grams of a gas A is introduced in an evacuated flask kept at 25 C; the pressure was found to be 1 atm .i.e.
Partial Pressure of gas A = 1 atm.
And let moles be nA = 2/Ma
if 3 g of another gas B is then added to the same flask;
Partial Pressure of gas B =1.5 atm –  1 atm. = 0.5 atm
moles of gas B  = 3/Mb
Again the temp and volume are constant therfore,
P1/n1 = P2/n2
hence,
1/(2/Ma) = 0.5/(2/Mb)
Ma/2 = 0.5Mb/2
Ma/2 = Mb/4
Ma = Mb/2
or
Ma/Mb = ½ => 1:2

Vikas is wrong his mistake is that,he has taken 2/mb in place of 3/mb...so dear learners please edit this.. All the best...

Guys answer will be 1:3 like. Partial pressure of gas A is 1 and total pressure is 1.5 so just apply formal of daltons law of partial pressure get no. Of moles of A and get molecular weight of A