Posted by **Arnold** on Thursday, March 18, 2010 at 12:27am.

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.

Hellpp pleasee!

- Chemistry -
**DrBob222**, Thursday, March 18, 2010 at 12:39am
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.

- Chemistry -
**ZAKES**, Sunday, August 26, 2012 at 10:07am
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

=0.700*37.6mmHg

=26.32mmHg

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