Antifreeze is made by mixing equal volumes of ethylene glycol (d=1.114 g/mL) and water. The resulting solution has a density of 1.040 g/mL. Calculate the antifreeze molarity, molality, mole fraction, mass percent, and volume percent.
Oh, antifreeze, the magical potion that keeps our engines from turning into icicles! Now, let's dive into the calculations and have some fun with numbers!
First, let's calculate the molarity:
To find the molarity, we need to figure out the number of moles of solute (ethylene glycol) per liter of solution.
Given that the density of the resulting solution is 1.040 g/mL, we can assume that 1 liter of the solution weighs 1040 g.
Since the density of ethylene glycol is 1.114 g/mL, the mass of the ethylene glycol in 1 liter of the solution can be calculated as:
mass of ethylene glycol = (density of ethylene glycol) * (volume of ethylene glycol) = 1.114 g/mL * 0.5 L = 0.557 g
Now, let's calculate the number of moles of ethylene glycol:
moles of ethylene glycol = (mass of ethylene glycol) / (molar mass of ethylene glycol)
The molar mass of ethylene glycol (C2H6O2) is approximately 62.07 g/mol.
moles of ethylene glycol = 0.557 g / 62.07 g/mol ≈ 0.00896 mol
Since the volume of the solution is 1 liter, the molarity can then be calculated by dividing the moles of solute by the volume of the solution in liters:
Molarity = (moles of ethylene glycol) / (volume of solution in liters) = 0.00896 mol / 1 L = 0.00896 M
Alright, now let's move on to molality:
Molality is the number of moles of solute per kilogram of solvent. Since water is considered the solvent, we need to figure out its mass in the solution.
mass of water = mass of solution - mass of ethylene glycol = 1040 g - 0.557 g ≈ 1039.443 g
Now, let's calculate the molality:
Molality = (moles of ethylene glycol) / (mass of water in kg) = 0.00896 mol / (1039.443 g / 1000 g/kg) ≈ 0.00864 mol/kg
Moving on to mole fraction:
The mole fraction of a component in a solution is the moles of that component divided by the total moles of all components.
moles of water = (mass of water) / (molar mass of water)
The molar mass of water (H2O) is approximately 18.02 g/mol.
moles of water = 1039.443 g / 18.02 g/mol ≈ 57.62 mol
Total moles of all components = moles of ethylene glycol + moles of water = 0.00896 mol + 57.62 mol ≈ 57.628 mol
Mole fraction of ethylene glycol = (moles of ethylene glycol) / (total moles of all components) = 0.00896 mol / 57.628 mol ≈ 0.0001557
Alright, let's march on to mass percent:
Mass percent is the mass of the solute divided by the total mass of the solution, multiplied by 100.
mass percent of ethylene glycol = (mass of ethylene glycol) / (mass of solution) * 100 = 0.557 g / 1040 g * 100 ≈ 0.0536%
Finally, let's finish up with volume percent:
Volume percent is the volume of the solute divided by the total volume of the solution, multiplied by 100.
volume of ethylene glycol = volume of solution = 0.5 L
volume percent of ethylene glycol = (volume of solute) / (volume of solution) * 100 = 0.5 L / 1 L * 100 = 50%
And there you have it! Now you know the antifreeze molarity (0.00896 M), molality (0.00864 mol/kg), mole fraction (0.0001557), mass percent (0.0536%), and volume percent (50%). Stay warm and safe out there!
To calculate the antifreeze molarity, molality, mole fraction, mass percent, and volume percent, we need to use the given information.
First, let's calculate the mass of the antifreeze solution by using the density formula:
Mass = Volume × Density
The volume of the solution is not explicitly mentioned, but we know that equal volumes of ethylene glycol and water are mixed. So, let's assume we have 100 mL of the solution.
Mass of the antifreeze solution = 100 mL × 1.040 g/mL = 104 g
Next, we need to determine the masses of ethylene glycol (EG) and water in the antifreeze solution.
Mass of EG = Volume of EG × Density of EG = 50 mL × 1.114 g/mL = 55.7 g
Mass of water = Volume of water × Density of water = 50 mL × 1.000 g/mL = 50 g
Now we can calculate the molarity, molality, mole fraction, mass percent, and volume percent.
1. Molarity (M):
Molarity = Number of moles of solute / Volume of solution in liters
The number of moles of EG can be calculated using its molar mass (62.07 g/mol):
Number of moles of EG = Mass of EG / Molar mass of EG = 55.7 g / 62.07 g/mol ≈ 0.897 mol
The volume of the solution is 100 mL = 0.1 L.
Molarity of antifreeze = Number of moles of EG / Volume of solution = 0.897 mol / 0.1 L ≈ 8.97 M
2. Molality (m):
Molality = Number of moles of solute / Mass of the solvent in kg
The mass of water is 50 g = 0.05 kg.
Molality of antifreeze = Number of moles of EG / Mass of water (solvent) = 0.897 mol / 0.05 kg = 17.94 mol/kg
3. Mole fraction:
Mole fraction of EG = Number of moles of EG / Total moles of substances in the solution
Total moles of substances in the solution = Number of moles of EG + Number of moles of water = 0.897 mol + 2.53 mol (from water)
Mole fraction of EG = 0.897 mol / (0.897 mol + 2.53 mol) = 0.262
Mole fraction of water = 2.53 mol / (0.897 mol + 2.53 mol) = 0.738
4. Mass percent:
Mass percent of EG = (Mass of EG / Mass of solution) × 100
Mass of the solution is 104 g (as calculated earlier).
Mass percent of EG = (55.7 g / 104 g) × 100 ≈ 53.56%
Mass percent of water = (50 g / 104 g) × 100 ≈ 48.08%
5. Volume percent:
Volume percent of EG = (Volume of EG / Volume of solution) × 100
Volume percent of EG = (50 mL / 100 mL) × 100 = 50%
Volume percent of water = (50 mL / 100 mL) × 100 = 50%
So, the antifreeze solution has:
- Molarity of approximately 8.97 M
- Molality of approximately 17.94 mol/kg
- Mole fraction of EG of approximately 0.262
- Mass percent of EG of approximately 53.56%
- Volume percent of EG of approximately 50%