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
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.
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