Problem: 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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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 componentMole 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


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Problem Details

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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