Part A If 0.300 mol of a nonvolatile nonelectrolyte are dissolved in 3.90 mol of
ID: 485181 • Letter: P
Question
Part A
If 0.300 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.
Part B
A solution is composed of 1.80 mol cyclohexane (Pcy=97.6 torr) and 2.90 mol acetone (Pac=229.5 torr). What is the total vapor pressure Ptotal above this solution?
Part B
(In solutions composed of two liquids (A and B), each liquid contributes to the total vapor pressure above the solution. The total vapor pressure is the sum of the partial pressures of the components: Ptotal=PA+PB=XAPA+XBPB) (where PA and PB are the vapor pressures of pure A and B, respectively.)
Explanation / Answer
A)
mole fraction of water,
X = mol of water / total mol
= 3.90 /(0.30 + 3.90)
= 0.9286
Now use:
P = Po*X
= 23.8 * 0.9286
= 22.1 torr
Answer: 22.1 torr
B)
Let A be cyclohexane and B be acetone
XA = 1.80 / (1.80 + 2.90)
= 0.383
XB = 1-XA =1-0.383 = 0.617
So,
P = PoA*XA + PoB*XB
= 97.6*0.383 + 229.5*0.617
= 179 torr
Answer: 179 torr