posted by Arnold .
The vapor pressures of ethanol (C2H5OH) and 1-propanol (C3H7OH) are 100 mmHg and 37.6 mmHg, respectively. Assume ideal behavior and calculate the partial pressures of ethanol and 1-propanol at 35 degrees Celsius over a solution of ethanol in 1-propanol, in which the mole fraction of ethanol is .300 .
I used the equation Pi = Xi * Ptotal. Where Pi is the partial pressure and Xi is the mole fraction.
So for ethanol I got Pi = .3 * 100mmHg = 30 mmHg (correct answer)
However for 1-propanol:
Pi = .3 * 37.6mmHg = 11.28 mmHg
But the book is telling me the answer is 26.3 mmHg.
And the book is right.
If the mole fraction for ethanol is 0.3, then the mole fraction for 1-propanol is 0.7 since mole fractions must add to 1.00.
THE MOLE FRACTION OF THE SOLUTES YIELD AT 1. THEREFORE IF ETHANOL HAS THE MOLE FRACTION OF 0.300, THE MOLE FRACTION OF 1-PROPANOL SHOULD BE 0.700 THEN USE THE EQUETION:Pi=Xi*P