the vapour pressure of ethanol (C2H5OH)and 1-propanol(C3H7OH)at35c are 100mmHg and 37.6mmHg,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.
yes
answer question
Pethanol = Xethanol*Poethanol
Pethanol = 0.3*100 = ?
Mole fractions always add to 1.0; therefore, mole fraction 1-propanol = 0.7, then P1-propanol = Xpropanol*Popropanol
Post your work if you get stuck.
To calculate the partial vapor pressures of ethanol and 1-propanol over a solution with a mole fraction of ethanol of 0.3 at a given temperature, we can use Raoult's law, which states that the vapor pressure of a component in a solution is equal to the mole fraction of that component multiplied by its pure vapor pressure.
1. Identify the given information:
- Vapor pressure of ethanol (C2H5OH) at 35°C: 100 mmHg
- Vapor pressure of 1-propanol (C3H7OH) at 35°C: 37.6 mmHg
- Mole fraction of ethanol: 0.3
2. Calculate the partial vapor pressure of ethanol:
The partial vapor pressure of ethanol (Pethanol) can be determined using Raoult's law:
Pethanol = Mole fraction of ethanol × Vapor pressure of ethanol
= 0.3 × 100 mmHg
= 30 mmHg
3. Calculate the mole fraction of 1-propanol:
The mole fraction of 1-propanol (M1propanol) can be calculated based on the given mole fraction of ethanol:
M1propanol = 1 - Mole fraction of ethanol
= 1 - 0.3
= 0.7
4. Calculate the partial vapor pressure of 1-propanol:
The partial vapor pressure of 1-propanol (P1propanol) can be determined using Raoult's law:
P1propanol = Mole fraction of 1-propanol × Vapor pressure of 1-propanol
= 0.7 × 37.6 mmHg
= 26.32 mmHg
Therefore, the partial vapor pressures of ethanol and 1-propanol over the solution, with a mole fraction of ethanol of 0.3, are approximately 30 mmHg and 26.32 mmHg, respectively.