Chemistry
posted by Arnold .
The vapor pressures of ethanol (C2H5OH) and 1propanol (C3H7OH) are 100 mmHg and 37.6 mmHg, respectively. Assume ideal behavior and calculate the partial pressures of ethanol and 1propanol at 35 degrees Celsius over a solution of ethanol in 1propanol, 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 1propanol:
Pi = .3 * 37.6mmHg = 11.28 mmHg
But the book is telling me the answer is 26.3 mmHg.
Hellpp pleasee!

And the book is right.
If the mole fraction for ethanol is 0.3, then the mole fraction for 1propanol 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 1PROPANOL SHOULD BE 0.700 THEN USE THE EQUETION:Pi=Xi*P
=0.700*37.6mmHg
=26.32mmHg 
26.32