Question

A) If 0.640 mol of a nonvolatile nonelectrolyte are dissolved in 3.60 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 .

B) A solution is composed of 1.60 mol cyclohexane (P∘cy=97.6 torr) and 2.00 mol acetone (P∘ac=229.5 torr). What is the total vapor pressure Ptotal above this solution?

Answer 1

A)

number of moles of non-volatile = 0.640 mole

number of moles of water = 3.60 mole

total number of moles of soltuion = 0.640+3.60 = 4.24 moles

mole fraction of water = number of moles of water/total number of moles of soltuion

Mole fractionof water = 3.60/4.24 = 0.849 moles

Vapour pressure of the solution = vapour pressure of water x mole fration of water

Vapour pressure ofthe solution = 23.8 x0.849 = 20.21 torr.

Vapour pressure of the solution = 20.21 torr

B)

number of moles of cyclohexane = 1.60 mol

number of moles of acetone = 2.00 mole

total number o fmoles = 1.60+2.00 = 3.60 moles

mole fraction of cyclohexane = 1.60/3.60 = 0.444

mole fraction of acetone = 2.00/3.60 = 0.556

Vapour pressure of cyclohexane = 97.6 x0.444 = 43.33 torr

vapour pressure of acetone = 229.5 x 0.556 = 127.6 torr

Total vapour pressure of the solution = 43.33 + 127.6 = 170.93 torr

Total vapour pressure of the solution = 170.93 torr