The pressure is found to be 1 atm when 2 g of a gaseous substance A are introduced into an initially evacuated flask kept at 25° C.3g of gaseous substance B are then added to the 2 g of A, and the pressure is now found to be 1.5 atm. Assuming ideal gas behavior, if the ratio of molecular weights, that is `M_A//M_B` is equal to `1//X`, find the value of X.

Answer

3

Explanation

`IXV=2/M_A R_T`
`1.5 xx V = (2/(MA)+3/(MB)) xx RT`
On diving
`1/(1.5)=(2/(MA))/((2/(MA)+3/(MB)))=1/(1+3/2+1/x)`