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johnny b
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Homework Statement
A solution of ethanol (eth) and chloroform (chl) at 45°C with xeth = 0.9900 has a vapor pressure of 177.95 torr. At this high dilution of chloroform, the solution can be assumed to be ideally dilute. The vapor pressure of pure ethanol at 45°C is 172.76 torr.
(a) Find the partial pressures of the gases in equilibrium with the solution
(b) Find the mole fractions in the vapor phase
(c) Find the Henry’s Law constant for chloroform in ethanol at 45°C
(d) Predict the vapor pressure and vapor-phase mole fractions at 45°C for a chloroform-ethanol solution with xeth = 0.9800. Compare with the experimental values P = 183.38 torr and =0.9242.
Homework Equations
partial pressure = total pressure x mole fraction
X(gas) = P / P(total)
Total vapor pressure = (X(eth)*P(eth)) + (X(chl)*P(chl))
K=P/x (henry's law constant)
The Attempt at a Solution
For part a, if I use the equation partial pressure = total pressure x mole fraction I get
p(eth)=0.99(177.95) --> p(eth)=171.0324 torr. x(chl)= 1-.99=.01
p(chl)=.01(177.95) --> (chl)=1.7276 torr.
is this right?
For part b assuming part a is right,
X(eth) = P(eth) / P(total)
x(eth)=(171.0324)/177.95 = 0.961
x(chl) 1-.961 = .039
part c:
Total vapor pressure = (X(eth)*P(eth)) + (X(chl)*P(chl))
177.95torr = (0.99*172.76) + (.01*P(chl))
P(chl)=691.76 torr = 0.910atm
K=P(chl)/x
K=0.910 atm/.01 = 91 atm <--- the x that I use is .01 right? not the .039 that I found in part b?
d:
This one I'm not sure about. The only thing I could think of is from the given part:
177.95 torr / .9900 = p / .9800
then p= 176.15 but this doesn't seem right and I don't know how to find the vapor-phase mole fractions
Sorry it's so much but I really want to make sure I'm doing this right. Any help is appreciated!