Chemistry

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!

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  1. 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.

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  2. 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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  3. 26.32

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  4. This question is very easy guys , it doesn't need teachers guide 😂😂 just look at the problem carefully the mole fraction of 1-propanol= 1-0.3 =0.7 (because the question says ........over the solution) hence we calculate mole fraction of 1-propanol depending on that of ethanol.

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