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.

100 g of liquid A (molar mass 140 g mol^(-1)) was dissolved in 1000 g

1.	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.
Answer» Solution :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.