🤓 Based on our data, we think this question is relevant for Professor Booker's class at University of Western Ontario.
A solution of 75.0 g of benzene (C6H6) and 75.0 g of toluene (C7H8) has a total vapor pressure of 80.9 mmHg at 303 K. Another solution of 100.0 g benzene and 50.0 g toluene has a total vapor pressure of 93.9 mmHg at this temperature.
Find the vapor pressure of pure benzene 303 K.
We’re being asked to determine the vapor pressure of pure benzene at 303 K. Recall that the vapor pressure of a solution can be given by Raoult’s Law.
The given solutions are composed of benzene and toluene, which are both volatile. For a solution with volatile solute and solvent, Raoult’s Law is given as:
where P˚ = vapor pressure of pure component and χ = mole fraction of component. Mole fraction is given by:
where i = van’t Hoff factor. We first need to determine the mole fraction of the solute and solvent for each solution. The molar mass of benzene, C6H6, is:
6 × 12.01 g/mol C = 72.06 g/mol C
6 × 1.01 g/mol H = 6.06 g/mol H
Molar Mass = 78.12 g/1 mol C6H6
The molar mass of toluene, C7H8, is:
7 × 12.01 g/mol C = 84.07 g/mol C
8 × 1.01 g/mol H = 8.08 g/mol H
Molar Mass = 92.15 g/1 mol C7H8