What is the equilibrium partial pressure of water vapor above a mixture of 37.5 g H2O and 62.5 g
HOCH2CH2OH at 55 degree C. The partial pressure of water at 55 degree C is 118.0 mm Hg. Assume ideal behavior for the solution.
To find the equilibrium partial pressure of water vapor above the mixture, we first need to calculate the mole fraction of water vapor in the solution.
1. Begin by calculating the moles of each component in the mixture:
- Moles of H2O: mass / molar mass
Moles of H2O = 37.5 g / 18 g/mol = 2.08 mol
- Moles of HOCH2CH2OH: mass / molar mass
Moles of HOCH2CH2OH = 62.5 g / (46 g/mol) = 1.36 mol
2. Calculate the total moles of the mixture:
Total moles = Moles of H2O + Moles of HOCH2CH2OH
= 2.08 mol + 1.36 mol = 3.44 mol
3. Calculate the mole fraction of water vapor (H2O) in the mixture:
Mole fraction of H2O = Moles of H2O / Total moles
= 2.08 mol / 3.44 mol = 0.6047
4. Now, we can calculate the partial pressure of water vapor using Raoult's law:
Partial pressure of water vapor = Mole fraction of H2O * Vapor pressure of pure water
= 0.6047 * 118.0 mm Hg
= 71.4 mm Hg
Therefore, the equilibrium partial pressure of water vapor above the mixture is approximately 71.4 mm Hg at 55 degrees Celsius.
To find the equilibrium partial pressure of water vapor above the mixture, you need to calculate the mole fraction of water vapor in the solution, and then use that mole fraction to determine the partial pressure of water vapor.
Step 1: Calculate the moles of each component:
Moles of H2O = mass / molar mass = 37.5 g / 18.015 g/mol = 2.08 mol
Moles of HOCH2CH2OH = mass / molar mass = 62.5 g / 62.07 g/mol = 1.01 mol
Step 2: Calculate the total moles of the solution:
Total moles = moles of H2O + moles of HOCH2CH2OH = 2.08 mol + 1.01 mol = 3.09 mol
Step 3: Calculate the mole fraction of water vapor:
Mole fraction of water vapor = moles of H2O / total moles = 2.08 mol / 3.09 mol = 0.672
Step 4: Calculate the equilibrium partial pressure of water vapor:
Equilibrium partial pressure = mole fraction of water vapor * total pressure
= 0.672 * 118.0 mm Hg = 79.30 mm Hg
Therefore, the equilibrium partial pressure of water vapor above the mixture is 79.30 mm Hg.
Find mols of each. The total moles.
mole fraction water = moles H2O/total mols.
pH2O = XH2O x Ponormal
Post your work if you get stuck.