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!
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
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
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.
Solution and answers this question
Andwer
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
26.32
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.
help me by answering
To calculate the partial pressure of 1-propanol in the solution, you correctly used the equation Pi = Xi * Ptotal. However, there seems to be an error in your calculation.
Let's go through the correct calculation step by step:
1. Given information:
- Vapor pressure of ethanol (C2H5OH) = 100 mmHg
- Vapor pressure of 1-propanol (C3H7OH) = 37.6 mmHg
- Mole fraction of ethanol (C2H5OH) = 0.300
2. Calculate the mole fraction of 1-propanol:
- Mole fraction of 1-propanol (C3H7OH) = 1 - mole fraction of ethanol = 1 - 0.300 = 0.700
3. Calculate the total pressure of the solution:
- Ptotal = Pethanol + P1-propanol
- Ptotal = (mole fraction of ethanol * vapor pressure of ethanol) + (mole fraction of 1-propanol * vapor pressure of 1-propanol)
- Ptotal = (0.300 * 100 mmHg) + (0.700 * 37.6 mmHg)
- Ptotal = 30 mmHg + 26.32 mmHg
- Ptotal = 56.32 mmHg
4. Calculate the partial pressure of 1-propanol using the correct equation:
- P1-propanol = mole fraction of 1-propanol * Ptotal
- P1-propanol = 0.700 * 56.32 mmHg
- P1-propanol = 39.424 mmHg
Therefore, the correct partial pressure of 1-propanol at 35 degrees Celsius in the solution with a mole fraction of ethanol of 0.300 is approximately 39.424 mmHg, not 11.28 mmHg as you calculated.
Please note that the values provided in the book may differ slightly due to rounding or using more precise data.