Part A If 0.880 mol of a nonvolatile nonelectrolyte are dissolved in 3.10 mol of
ID: 525746 • Letter: P
Question
Part A
If 0.880 mol of a nonvolatile nonelectrolyte are dissolved in 3.10 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.
SubmitHintsMy AnswersGive UpReview Part
Correct
For solutions that contain solutes that dissociate, the number of moles of particles produced must be taken into consideration in determining the mole fraction of the solvent. For solutions that contain solutes that are volatile, the vapor pressure of the solution will be determined by both the solvent and the solutes.
Solutions containing volatile solutes
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.
Part B
A solution is composed of 1.90 mol cyclohexane (Pcy=97.6 torr) and 2.30 mol acetone (Pac=229.5 torr). What is the total vapor pressure Ptotal above this solution?
Express your answer with the appropriate units.
SubmitHintsMy AnswersGive UpReview Part
Correct
If a third volatile component had been present in this solution, the total pressure would be calculated in a similar manner but would involve three terms rather than two. For example,
Ptotal=PA+PB+PC=XAPA+XBPB+XCPC
Significant Figures Feedback: Your answer 169.8torr was either rounded differently or used a different number of significant figures than required for this part.
Part C
As you saw in Part B, the vapor above the cyclohexane-acetone solution is composed of both cyclohexane vapor and acetone vapor. What mole fraction of the vapor above the solution, Xcy(vapor), is cyclohexane?
Express your answer numerically.
Xcy(vapor)=
I only need part C
PH2O = 18.5 torrExplanation / Answer
Ans Part C
While calculating the vapour pressure above the solution , you might have found the indiviadual contributions of each cyclohexane and acetone and then added them
Mole fraction of cyclohexane = 1.90 / ( 1.90 + 2.30) = 0.45
pressure exerted = 0.45 x 97.6 = 44.15 torr
now the total vapour pressure is 169.8 torr
So the mole fraction of vapour pressure of cyclohexane = 44.15 / 169.8
=0.26
so 0.26 mole fraction of vapour above the solution is cyclohexane .