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