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 .

Question

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!

Answer 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.

Answer 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

Answer 3

26.32

Answer 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.

Answer 5

The vapour pressure of ethanol (C2
H5
OH) and 1-propanol (C3
H7
OH) at 35°C are
100 mmHg and 37.6 mmHg, respectively. Assuming ideal behaviour, calculate the
partial vapour pressures of ethanol and 1-propanol over a solution, in which the
mole fraction of ethanol is 0.3

Answer 6

Solution and answers this question

Answer 7

Andwer

Answer 8

Calculate the freezing point of a solution of 3.46g of Compound X,in160g of benzen ,when a separate sample of X was Vaporised it density was found to be 3.77g/l at 116°Cand 773torr, The freezing point of pure benzen is 5.45°C and Kf is 5.12°C kg /mol

Answer 9

help me by answering