100 g of liquid A (molar mass 140 g `mol^(-1)`) was dissolved in 1000 g of liuquid B (molar mass 180 g `mol^(-1)`). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Explanation

Number of moles of liquid A :
`n_(A)=(100)/(140)=0.7143` mol
Number of moles of liquid B :
`n_(B)=(1000)/(180)=5.556` mol
Then mole fraction of A,
`x_(A)=(n_(A))/(n_(A)+n_(B))`
`= (0.714)/(0.714+5.556)=0.114`
And, mole fraction of B,
`= x_(B)=1-0.114`
= 0.8861
Vapour pressure of pure liquid B,
`p_(B)^(0)=500` torr
Therefore, vapour presure of liquid B in the solution,
`p_(B)=p_(B)^(0)x_(B)=500xx0.886`
= 443 torr
Total vapour pressure of the solution of liquid A in hte solution,
`p_(A)=p_("total")-p_(B)`
`= 475-433`
= 32 torr
Now, `p_(A)^(0)x_(A)`
`p_(A)^(0)=(p_(A))/(x_(A))`
`= (32)/(0.114)=280.5` torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.