Part A If 0.440 mol of a nonvolatile nonelectrolyte are dissolved in 3.40 mol of
ID: 1002881 • Letter: P
Question
Part A
If 0.440 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 .
Express your answer with the appropriate units.
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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 1.50 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 =Explanation / Answer
total no of moles = 0.44+ 3.4 = 3.84 moles
mole fraction of water = no of moles of water/ total no of moles
= 3.4/3.84 = 0.885
vapor pressure of solution from raout's law P = molefraction * water vapor pressure
= 0.885*23.8 = 21.063 torr
no fo moles of cyclohexane ncy = 1.5 moles
no of moles of acetone nac = 2.4 mole
total no of moles = ncy+ nac = 1.5+2.4 = 3.9
mole fraction of cyclohexane Xcy = ncy/ncy+nac
= 1.5/1.5+2.4 = 0.384
mole fraction of acetone Xac = nac/ncy+ nac
= 2.4/1.5+2.4 = 0.615
vapor pressure of cyclohexane = mole fraction *partial pressure
=0.384*97.6 = 37.4784 torr
vapor pressure of acetone = mole fraction *partial pressure
= 0.615*229.5 = 141.1425 torr
total pressure = 37.4784 + 141.1425 = 178.6209 torr