An ideal solution is prepared by mixing 234 g of benzene ( C ₆ H ₆ ) and 92 g of toluene ( C ₇ H ₈ ) at a certain temperature. If the vapour pressures of pure benzene and pure toluene at this temperature are 120 mm Hg and 40 mm Hg respectively, the total vapour pressure of the solution is:

Step-by-step solution

Molar mass of benzene ( C ₆ H ₆ ) = 78 g mol ⁻¹ Molar mass of toluene ( C ₇ H ₈ ) = 92 g mol ⁻¹ Number of moles of benzene, n_B = 234 78 = 3 mol Number of moles of toluene, n_T = 92 92 = 1 mol Total number of moles = 3 + 1 = 4 mol Mole fraction of benzene, X_B = 3 4 Mole fraction of toluene, X_T = 1 4 According to Raoult's law, the total vapour pressure is: P_ total = X_B P_B^ + X_T P_T^ P_ total = ( 3 4 120 ) + ( 1 4 40 ) P_ total = 90 + 10 = 100 mm Hg Answer: 100 mm Hg