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.