A solution contains 50.0 g of heptane (C7H16) and 50.0 g of octane (C8H18) at 25
ID: 477827 • Letter: A
Question
A solution contains 50.0 g of heptane (C7H16) and 50.0 g of octane (C8H18) at 25oC. The vapor pressures of pure heptane and pure octane at 25oC are 45.8 torr and 10.9 torr, respectively. Assuming ideal behavior, answer the following: a. What is the vapor pressure of each of the solution components in the mixture? b. What is the total pressure above the solution? c. What is the composition of each vapor in mass percent? d. Why is the composition of the vapor different from the composition of the solution? e. If the vapor pressure above the solution is measured to be 28.2 torr, what does that tell you about the solute and solvent interactions?
Explanation / Answer
a) According to Raoult's law vapor pressure of each component = Po x its mole fraction
moles of heptane = mass /molar mass
= 50g/100 = 0.5
moles of octane = 50/114 = 0.4385
mole fraction of heptane = 0.5/(0.5+0.4385)
=0.5327
mole fraction of octane = 0.4385/0.9385
= 0.4672
Thus vapor of pressure of heptane = 45.8x0.5327
=24.39 torr
and vapor pressure of octane =10.9 x0.4672
=5.09 torr
b) total vapor pressure above solution = vapor pressure of heptane + vapor pressure of octane
=24.39 +5.09
=29.48 torr
c)Composition of vapor
mole fraction of heptane in vapor = vapor pressure of heptane /total pressure (Dalton's law)
= 24.39/29.48
=0.8273
and mole fracion of octane in vapor = 1-0.8273
=0.1727
d) The composition of the solution and composition of vapor are different because, more volatile component goes into vapor more. Since heptane is more volatile(higher vapor pressure) it has more volatile nature and thus it is more in vapor state.
e) If the solution is ideal it shows vapor pressure =29.48 torr, but if it has 28.2 torr, tha tis less than the calculated value by Raoult's law, then it is showing negative deviation from ideal behaviour.
The solvent-solute interactions are stronger than solute-solute or solvent-solvent interactions.
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