Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
The mole fraction of benzene in vapour phase is 0.599 (or approximately 0.60).
Explanation
This problem involves applying Raoult's law for ideal solutions to find the vapour phase composition. First, we calculate the moles and mole fractions in the liquid phase. Then, using Raoult's law, we determine the partial pressures of each component. Finally, using Dalton's law of partial pressures, we find the mole fraction in the vapour phase.
Solution Steps
Step 1: Calculate moles of benzene and toluene Molar mass of benzene (C₆H₆) = 78 g mol⁻¹ Molar mass of toluene (C₇H₈) = 92 g mol⁻¹ Moles of benzene = 80 g / 78 g mol⁻¹ = 1.026 mol Moles of toluene = 100 g / 92 g mol⁻¹ = 1.087 mol
Step 2: Calculate mole fractions in liquid phase Total moles = 1.026 + 1.087 = 2.113 mol Mole fraction of benzene (x_benzene) = 1.026 / 2.113 = 0.486 Mole fraction of toluene (x_toluene) = 1.087 / 2.113 = 0.514
Step 3: Apply Raoult's law to find partial pressures For ideal solution: p = x × p⁰ Partial pressure of benzene = mm Hg = 24.65 mm Hg Partial pressure of toluene = mm Hg = 16.48 mm Hg
Step 4: Calculate total vapour pressure p_total = p_benzene + p_toluene = 24.65 + 16.48 = 41.13 mm Hg
Step 5: Calculate mole fraction of benzene in vapour phase Using Dalton's law: p_benzene = y_benzene × p_total y_benzene = p_benzene / p_total = 24.65 / 41.13 = 0.599