At 58.8 degrees C and at a total pressure of 1.00 atm the mole percent of acetone in the vapor state above a solution of acetone and water containing 70. mol % acetone is 87.5%. Assuming the solution to obey Raoult's Law, determine the vapor pressure of pure water at this temperature. Comparing your results to the actual value of 141 mmHg, what does this suggest about the acetone-water solution?

Question

At 58.8 degrees C and at a total pressure of 1.00 atm the mole percent of acetone in the vapor state above a solution of acetone and water containing 70. mol % acetone is 87.5%. Assuming the solution to obey Raoult's Law, determine the vapor pressure of pure water at this temperature. Comparing your results to the actual value of 141 mmHg, what does this suggest about the acetone-water solution?

I calculated the first part like so:
1.00 atm= .875 * x
1.14 atm= x

I converted 141 mmHg to atm, which turned out to be .186- obviously quite a bit lower than the answer found above. I'm not sure what this says about the mixture, however.

I get a smaller number than 1.14.
Isn't
P_(water)/P_(total) = X_(water) = 0.875.

Then P_(total)*X_(water) = P_(water) = 1*0.875 = 0.875 atm. That's still very much higher than 0.186 atm. Would you think it says the acetone/water mixture is not an ideal solution?
Check my thinking. Check my work.

Answer 1

To solve this problem, we first need to understand Raoult's Law. According to Raoult's Law, the partial pressure of a component in the vapor phase above a solution is directly proportional to its mole fraction in the liquid phase. In this case, we are considering a solution of acetone and water.

Let's denote the partial pressure of acetone in the vapor phase as P(acetone) and the mole fraction of acetone in the liquid phase as X(acetone). According to Raoult's Law, we can write:

P(acetone)/P(total) = X(acetone)

Given that the mole percent of acetone in the vapor state is 87.5% and the total pressure is 1.00 atm, we can substitute these values into the equation:

P(acetone)/1.00 atm = 0.875

Solving for P(acetone), we find P(acetone) = 0.875 atm.

To determine the vapor pressure of pure water at this temperature, we can subtract the pressure of acetone from the total pressure:

P(water) = P(total) - P(acetone)
P(water) = 1.00 atm - 0.875 atm
P(water) = 0.125 atm

Now, let's compare this to the actual value of 141 mmHg. To convert mmHg to atm, we divide by 760 mmHg/atm:

P(water) = (141 mmHg) / (760 mmHg/atm)
P(water) ≈ 0.1855 atm

Comparing the calculated value of 0.125 atm to the actual value of 0.1855 atm, we can see that the calculated value is lower. This suggests that the acetone-water solution deviates from Raoult's Law and is not an ideal solution.

Therefore, based on your calculations, it suggests that the acetone-water solution is not an ideal solution and that there are deviations from Raoult's Law.