two liquids having vapour pressure P1 and P2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2 . the ratio of their moles in vapour state would be

1) 1 :1

2) 1 :2

3) 2:1

4) 3:2

Vapour pressure of liquid 1, P1 = 2units
Vapour pressure of liquid 2, P2 = 1unit
Moles of liquid 1 = 1
Moles of liquid 2 = 2
Mole fraction of liquid 1  = 1/3
Mole fraction of liquid 2 = 2/3
Using Raoult's law:
Partial pressure of liquid 1 = Vapour pressure of liquid 1 $\times$ mole freaction of liquid 1
Partial pressure of liquid 1 = 2 $\times$ (1/3) = 2/3

Partial pressure of liquid 2 = ​Vapour pressure of liquid 2 $\times$ mole freaction of liquid 2
Partial pressure of liquid 2 = 1​$\times$(2/3) = 2/3

Total pressure p = partial pressure of 1 + partial pressure of 2 = $\frac{2}{3}+\frac{2}{3} = \frac{4}{3}$

Mole fraction of liquid 1 in vapour phase = $\frac{partial pressure of liquid 1}{total vapour pressure} = \frac{(2/3)}{(4/3)} = \frac{1}{2}$

Mole fraction of liquid 2 in vapour phase = $\frac{partial pressure of liquid 2}{total vapour pressure} = \frac{(2/3)}{(4/3)} = \frac{1}{2}$

Ratio of moles = 1:1

Hence correct option is (1).