An ideal solution is prepared by mixing 64 g of methanol ( CH ₃ OH ) and 138 g of ethanol ( C ₂ H ₅ OH ) at a constant temperature. If the vapour pressures of pure methanol and pure ethanol at this temperature are 90 mm Hg and 40 mm Hg respectively, what is the total vapour pressure of the solution?

Step-by-step solution

Molar mass of methanol ( CH ₃ OH ) = 12 + 4 + 16 = 32 g mol ⁻¹ Moles of methanol = 64 32 = 2 mol Molar mass of ethanol ( C ₂ H ₅ OH ) = 24 + 6 + 16 = 46 g mol ⁻¹ Moles of ethanol = 138 46 = 3 mol Total number of moles = 2 + 3 = 5 mol Mole fraction of methanol, X_ CH ₃ OH = 2 5 = 0.4 Mole fraction of ethanol, X_ C ₂ H ₅ OH = 3 5 = 0.6 Using Raoult's law, the total vapour pressure of the solution is given by: P_ total = P^ _ CH ₃ OH X_ CH ₃ OH + P^ _ C ₂ H ₅ OH X_ C ₂ H ₅ OH P_ total = (90 0.4) + (40 0.6) P_ total =