Two volatile liquide A and B are mixed in mole ratio 1:3, to form an ideal liquid solution Vapour pressure of pure. A and B are 600 torr and 200 torr. Volume of container for vapour is 76 litre. `(R= 1//12" " "litre" " "atm//K//"mole")`
Calculate the mole of A in vapour phase at 300 K, assuming negligible amount of vapour is present compared to liquid in container

Options

A. `2` mole
B. `0.3` mole
C. `1.2` mole
D. `0.6` mole

Answer

D

Explanation

`P_(T) = 600xx(1)/(4)+200xx(3)/(4) = (1200)/(4) = 300` torr
`n_(T)=((PV)/(RT)) = ((300)/(760)xx76)/((1)/(12)xx300) = 1.2`mole
`Y_(A) = (P_(A)^(0)X_(A))/(P_(T)) = (600xx(1)/(4))/(300) = 0.5`
`n_(A)` in vapour `= 0.5 xx 1.2 = 0.6` mole