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Question

Calculate the vapor pressure at 85.0°C of a solution prepared by dissolving 0.600 mol of liquid dibromoethane (C₂H₄Br₂, P^o=127 torr) in 1.80 mol of liquid dibromopropane (C₃H₆Br₂, P^o= 173 torr).

Can someone help me understand how to do this?

Answer 1

Read up on Dalton's law of partial pressure and Raoult's law.

When a mixture is formed, the vapor pressure of the pure solvent is reduced by the addition of the solute. Sometimes the solute is non volatile and sometimes it is volatile. In the present problem, both are volatile, so the vapor pressure of each liquid will be reduced. The amount it is reduced is related to the mole fraction.

So, let's first determine the mole fraction of each liquid.

moles C2H4Br2 = 0.600 moles

moles C3H6Br2 = 1.80 moles

Total moles = 0.600 + 1.80 = 2.40 moles

mole fraction C2H4Br2 = 0.600 mol / 2.40 mol = 0.25

mole fraction C3H6Br2 = 1.80 / 2.40 = 0.75

Vapor pressure of C2H4Br2 = mole fraction x vapor pressure of pure = 0.25 x 127 torr = 31.8 torr

Vapor pressure of C3H6Br2 = mole fraction x vapor pressure of pure = 0.75 x 173 torr = 130 torr

Total vapor pressure of the solution = 32 torr + 130 torr = 162 torr