Ethanol and methanol form an ideal solution when mixed. What is the vapor pressure of a solution prepared by mixing equal masses of ethanol and methanol? (The vapor pressures of ethanol and methanol are 44.5 mmHg and 88.7 mmHg, respectively.)
I'm not sure what to do. Any help will be great!
I'm not sure what to do. Any help will be great!
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Hi Cassio!
This question requires you to use Raoult's law to figure out the change in vapor pressure of each substance once they're mixed. Once you do that, you can then add the vapor pressures together to get the total vapor pressure of the solution.
Raoult's law, in short, basically states that whenever a solute is added to a solvent, the solvent's vapor pressure will decrease. This can be represented by the equation: P = A times X where P represents the partial pressure of a component, A represents the vapor pressure of the pure component and X represents the mole fraction of the component. We must use this equation twice to find the new vapor pressures of ethanol and methanol before we add them together.
Before we use the equation, we must find the mole fraction of ethanol and methanol. The question never specified a definite mass; however, it did state that equal masses were mixed and so we can assume any mass for each of the two substances as long as they're equal. I will arbitrarily use 5.0 grams as the mass of each substance being added for this example. Now convert each mass into moles: Ethanol, (5.0g)(1 mole/46g) = 0.11 mols. Methanol, (5.0g)(1 mole/32g) = 0.16 mols. Now that we know the moles of each substance, we can then form a mole ratio to be placed into our vapor pressure equation. Take note that the total moles is then 0.27 (0.11 +0.16) and you simply divide the moles of either ethanol or methanol over 0.27 to find its mole ratio.
Ethanol: P = (44.5 mmHg)(0.11/0.27) = 18 mmHg
Methanol: P = (88.7 mmHg)(0.16/0.27) = 53 mmHg
Therefore, the vapor pressure of the solution is 18 + 52 = 71 mmHg. I used significant figures throughout my entire calculation so don't worry about it!
I hope this helped and good luck with chemistry :)
This question requires you to use Raoult's law to figure out the change in vapor pressure of each substance once they're mixed. Once you do that, you can then add the vapor pressures together to get the total vapor pressure of the solution.
Raoult's law, in short, basically states that whenever a solute is added to a solvent, the solvent's vapor pressure will decrease. This can be represented by the equation: P = A times X where P represents the partial pressure of a component, A represents the vapor pressure of the pure component and X represents the mole fraction of the component. We must use this equation twice to find the new vapor pressures of ethanol and methanol before we add them together.
Before we use the equation, we must find the mole fraction of ethanol and methanol. The question never specified a definite mass; however, it did state that equal masses were mixed and so we can assume any mass for each of the two substances as long as they're equal. I will arbitrarily use 5.0 grams as the mass of each substance being added for this example. Now convert each mass into moles: Ethanol, (5.0g)(1 mole/46g) = 0.11 mols. Methanol, (5.0g)(1 mole/32g) = 0.16 mols. Now that we know the moles of each substance, we can then form a mole ratio to be placed into our vapor pressure equation. Take note that the total moles is then 0.27 (0.11 +0.16) and you simply divide the moles of either ethanol or methanol over 0.27 to find its mole ratio.
Ethanol: P = (44.5 mmHg)(0.11/0.27) = 18 mmHg
Methanol: P = (88.7 mmHg)(0.16/0.27) = 53 mmHg
Therefore, the vapor pressure of the solution is 18 + 52 = 71 mmHg. I used significant figures throughout my entire calculation so don't worry about it!
I hope this helped and good luck with chemistry :)