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.
The vapour pressure of pure liquid A is 280 torr and its vapour pressure in the solution is 32 torr.
Explanation
This problem applies Raoult's law for binary solutions of volatile liquids. The key steps involve calculating mole fractions of both components, then using the relationship that total vapour pressure equals the sum of partial pressures. Since we know the total vapour pressure and can calculate the partial pressure of B using its mole fraction and known pure vapour pressure, we can find the partial pressure of A by subtraction. Finally, using Raoult's law (PA = xA × P°A), we can determine the vapour pressure of pure A.
Solution Steps
Step 1: Calculate moles of A and B Moles of A = 100 g / 140 g mol⁻¹ = 0.714 mol Moles of B = 1000 g / 180 g mol⁻¹ = 5.556 mol
Step 2: Calculate mole fractions Total moles = 0.714 + 5.556 = 6.27 mol Mole fraction of A (xA) = 0.714 / 6.27 = 0.114 Mole fraction of B (xB) = 5.556 / 6.27 = 0.886
Step 3: Calculate partial vapour pressure of B According to Raoult's law: PB = xB × P°B PB = torr = 443 torr
Step 4: Calculate partial vapour pressure of A in solution PA = Ptotal - PB = 475 - 443 = 32 torr
Step 5: Calculate vapour pressure of pure A Using Raoult's law: PA = xA × P°A 32 = 0.114 × P°A P°A = 32 / 0.114 = 280 torr