Question

Part A

If 0.560 mol  of a nonvolatile nonelectrolyte are dissolved in 3.40 mol of water, what is the vapor pressure PH2O of the resulting solution? The vapor pressure of pure water is 23.8 torr at 25 ∘C .

Part B

A solution is composed of 1.90 mol cyclohexane (P∘cy=97.6 torr) and 2.80 mol acetone (P∘ac=229.5 torr). What is the total vapor pressure Ptotal above this solution?

Answer 1

Part A

Number of moles of nonvolatile nonelectrolyte = 0.560 mol

Number of moles of water = 3.40 moles

Mole fraction of water = 3.40/(3.40+0.560) = 0.8585

Vapor Pressure of the resulting solution = mole fraction of water * Vapor Pressure of pure water

=> 0.8585 * 23.8

=> 20.434 torr

Part B

Number of moles of Cyclohexane = 1.90 moles

Number of moles of acetone = 2.80 moles

Mole fraction of Cyclohexane = 1.90/(1.90+2.80) = 0.4043

Mole fraction of acetone = 1 - Mole fraction of Cyclohexane = 0.5957

Total Vapor Pressure = mole freaction of cyclohexane * vapor pressure of pure cyclohexane + mole freaction of acetone * vapor pressure of pure acetone

=> 0.4043 * 97.6 + 0.5957 * 229.5

=> 176.17 torr