Part A If 0.120 mol of a nonvolatile nonelectrolyte are dissolved in 3.90 mol of
ID: 990859 • Letter: P
Question
Part A
If 0.120 mol of a nonvolatile nonelectrolyte are dissolved in 3.90 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 .
Express your answer with the appropriate units.
PH2O =
Part B
A solution is composed of 1.30 mol cyclohexane (Pcy=97.6 torr) and 2.40 molacetone (Pac=229.5 torr). What is the total vapor pressure Ptotal above this solution?
Express your answer with the appropriate units.
PH2O =
Part B
A solution is composed of 1.30 mol cyclohexane (Pcy=97.6 torr) and 2.40 molacetone (Pac=229.5 torr). What is the total vapor pressure Ptotal above this solution?
Express your answer with the appropriate units.
Ptotal =Explanation / Answer
Part A
From Raoult's Law:
P = (P0a) x (Xa)where P0 is the vapor pressure of pure a and Xa is the mole fraction of a.
The mole fraction of water in the solution is:
3.90 mol / (3.90 + 0.12) mol = 3.90 / 4.02 = 0.97
The vapor pressure of the solution, by Raoult's Law is:
P = 0.97 X 23.8 torr = 23.086 torr
Part B
Ptotal = (P0a x Xa) + (P0b x Xb)
mole fraction cyclohexane = (1.30) / (1.30 + 2.40) = 1.30 / 3.70 = 0.35
mole fraction acetone = 1 - 0.35 = 0.65
Ptotal = (97.6 torr)(0.35) + (229.5 torr)(0.65)
= 34.16 torr + 149.18 torr
= 183.34 torr