An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity, and mole fraction of the ethylene glycol.
40% means 40 g ethylene glycol/100 g solution.
40 g to moles = 40/molar mass.
Use density to convert 100 g soln to volume, then M = moles/L.
Convert % to m
40g/100 g = 40g glycol/(40g glycol + 60 g water).
Convert 40 g glycol to moles.
Convert 60 g water to kg.
m = moles/kg solvent.
For mole fraction starting with 40% glycol.
moles glycol = 40/molar mass
moles solvent = 60 g/molar mass
mole fraction glycol = moles glycol/total moles.
To find the molality of the ethylene glycol in the solution, we need to calculate the moles of ethylene glycol present per kilogram of solvent (water).
Step 1: Calculate the mass of the solution
Given that the density of the solution is 1.05 g/cm^3, we can assume 1 cm^3 of the solution has a mass of 1.05 g.
Therefore, the mass of 1 liter (1000 cm^3) of the solution is 1.05 kg.
Step 2: Calculate the mass of ethylene glycol in the solution
Since the solution is 40.0% ethylene glycol by mass, the mass of ethylene glycol in 1 liter of the solution is 40.0% of 1.05 kg.
Mass of ethylene glycol = 40.0% x 1.05 kg = 0.40 x 1.05 kg = 0.42 kg
Step 3: Calculate the moles of ethylene glycol
The molar mass of ethylene glycol (C2H6O2) is:
(2 x 12.01 g/mol) + (6 x 1.01 g/mol) + (2 x 16.00 g/mol) = 62.08 g/mol
Moles of ethylene glycol = Mass of ethylene glycol / Molar mass of ethylene glycol
Moles of ethylene glycol = 0.42 kg / 62.08 g/mol = 0.00676 mol
Step 4: Calculate the molality
Molality (m) = Moles of solute / Mass of the solvent in kg
The mass of the solvent (water) is the mass of the solution minus the mass of the solute (ethylene glycol).
Mass of solvent = Mass of solution - Mass of solute
Mass of solvent = 1.05 kg - 0.42 kg = 0.63 kg
Molality (m) = 0.00676 mol / 0.63 kg = 0.0107 mol/kg
Therefore, the molality of the ethylene glycol in the solution is 0.0107 mol/kg.
To find the molarity and mole fraction of the ethylene glycol, we need to know the volume of the solution.
To calculate the molality, molarity, and mole fraction of the ethylene glycol in the aqueous antifreeze solution, we need to use the given information about the mass percentage and density.
1. Molality (m):
Molality is defined as the number of moles of solute (in this case, ethylene glycol) per kilogram of solvent (water). To calculate the molality, we need to find the number of moles of ethylene glycol and the mass of water in kilograms.
First, let's assume we have 100 g of the aqueous solution. Since the mass percentage of ethylene glycol is given as 40.0%, we can calculate the mass of ethylene glycol in the solution:
Mass of ethylene glycol = 40.0% * 100 g = 40 g
The remaining mass is water:
Mass of water = 100 g - 40 g = 60 g
To convert the mass of water to kilograms (since molality is defined in terms of kilograms of water):
Mass of water in kg = 60 g / 1000 = 0.06 kg
Now, we need to calculate the number of moles of ethylene glycol. To do this, we'll use the molar mass of ethylene glycol, which is 62.07 g/mol:
Number of moles of ethylene glycol = Mass of ethylene glycol / Molar mass of ethylene glycol
Number of moles of ethylene glycol = 40 g / 62.07 g/mol ≈ 0.644 mol
Finally, we can calculate the molality:
Molality (m) = Number of moles of ethylene glycol / Mass of water in kg
Molality (m) = 0.644 mol / 0.06 kg ≈ 10.73 mol/kg
2. Molarity (M):
Molarity is defined as the number of moles of solute (ethylene glycol) per liter of solution. To calculate the molarity, we need to find the number of moles of ethylene glycol and the volume of the solution in liters.
Since the density of the solution is given as 1.05 g/cm3, we can calculate the volume of the 100 g solution (since 100 g of the solution contains 40 g of ethylene glycol):
Volume of solution = Mass of solution / Density
Volume of solution = 100 g / 1.05 g/cm3 = 95.24 cm3
However, we need to convert cm3 to liters:
Volume of solution in liters = 95.24 cm3 * (1 L / 1000 cm3) = 0.09524 L
Now, we can calculate the molarity:
Molarity (M) = Number of moles of ethylene glycol / Volume of solution in liters
Molarity (M) = 0.644 mol / 0.09524 L ≈ 6.76 mol/L
3. Mole fraction (χ):
Mole fraction is defined as the ratio of the number of moles of a component (ethylene glycol) to the total number of moles of all components in the mixture.
To calculate the mole fraction, we'll use the number of moles of ethylene glycol and the number of moles of water.
Number of moles of water = Mass of water / Molar mass of water
Number of moles of water = 60 g / 18.02 g/mol ≈ 3.33 mol
The total number of moles in the solution = Number of moles of ethylene glycol + Number of moles of water
The total number of moles in the solution = 0.644 mol + 3.33 mol ≈ 3.974 mol
Finally, we can calculate the mole fraction of ethylene glycol:
Mole fraction of ethylene glycol (χ) = Number of moles of ethylene glycol / Total number of moles in the solution
Mole fraction of ethylene glycol (χ) = 0.644 mol / 3.974 mol ≈ 0.162