How would you prepare 500.0 of an aqueous solution of glycerol with a vapor pressure of 24.8mm Hg at 26 degree Celsius (vapor pressure of pure water=25.21mm Hg)? assume the density of the solution to be 1.00 g/mL.
500.0 what? I'll assume you meant 500.0 mL.
Psoln = XH2O*Po
Substitute and solve for XH2O = mole fraction H2O to produce vapor pressure of 24.80 mm.
If we let x = grams glycerol, then 500-x = grams H2O
mols glycerol = x/92.09
mols H2O = (500-x)/18.015
Plug into mole fraction H2O
nH2O/(nH2O + nglycerol) = XH2O and solve for g glycerol and g H2O.
To prepare a 500.0 mL aqueous solution of glycerol with a specific vapor pressure, you need to follow these steps:
Step 1: Determine the vapor pressure difference between the desired solution and pure water.
Given:
- Vapor pressure of the solution (glycerol): 24.8 mm Hg
- Vapor pressure of pure water: 25.21 mm Hg
Subtract the vapor pressure of pure water from the desired vapor pressure:
Vapor pressure difference = 24.8 mm Hg - 25.21 mm Hg = -0.41 mm Hg
Step 2: Calculate the mass of glycerol needed to achieve the desired vapor pressure difference.
Given:
- Density of the solution: 1.00 g/mL
- Volume of the solution: 500.0 mL
Density = Mass / Volume
Rearranging the equation, Mass = Density * Volume
Mass = 1.00 g/mL * 500.0 mL = 500.0 g
Step 3: Calculate the mole fraction of glycerol in the solution.
Mole fraction (ϕ) is defined as the moles of glycerol divided by the total moles of the solution.
Mole fraction (ϕ) = Moles of Glycerol / Total Moles of Solution
To find the moles of glycerol, divide the mass of glycerol by its molar mass.
Given:
- Molar mass of glycerol (C3H8O3) = 92.09 g/mol
- Mass of glycerol = 500.0 g
Moles of Glycerol = Mass of Glycerol / Molar mass of Glycerol
Moles of Glycerol = 500.0 g / 92.09 g/mol ≈ 5.43 mol
To find the total moles of the solution, divide the mass of the solution by its molar mass.
Given:
- Molar mass of H2O = 18.015 g/mol
- Density of the solution = 1.00 g/mL
- Volume of the solution = 500.0 mL
Mass of the solution = Density of the solution * Volume of the solution
Mass of the solution = 1.00 g/mL * 500.0 mL = 500.0 g
Moles of H2O = Mass of H2O / Molar mass of H2O
Moles of H2O = 500.0 g / 18.015 g/mol ≈ 27.75 mol
Total Moles of Solution = Moles of Glycerol + Moles of H2O
Total Moles of Solution = 5.43 mol + 27.75 mol ≈ 33.18 mol
Mole fraction (ϕ) = Moles of Glycerol / Total Moles of Solution
Mole fraction (ϕ) = 5.43 mol / 33.18 mol ≈ 0.164
So, the mole fraction of glycerol in the solution is approximately 0.164.
In summary,
To prepare a 500.0 mL aqueous solution of glycerol with a vapor pressure of 24.8 mm Hg at 26 degrees Celsius, you would need approximately 500.0 grams of glycerol. The mole fraction of the glycerol in the solution would be approximately 0.164.