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


Given-


Mass of liquid A, WA = 100g, Molar mass, MA = 140 g mol-1


Mass of liquid B, WB = 1000 g, Molar mass, MB = 180 g mol-1


Using the formula below calculate the no. of moles in liquid A and B.



Number of moles of liquid A, MA = 100/140 = 0.714 mol-1


Number of moles of liquid B, MB = 1000/ 180 = 5.556 mol-1


Using the formula,


,


we calculate the mole fraction of liquids A and B.


Mole fraction of A,



Mole fraction of B,



Vapour pressure of pure liquid B, PoB = 500 torr (given)


According to Henry`s law,





Having given the total vapour pressure of the solution, Ptotal = 475 torr,






Using Henry`s law again to get the vapour pressure of pure liquid A,





Thus, the vapour pressure of pure liquid A = 280.7 torr and


vapour pressure of liquid A in the solution = 32 torr


36
1