An aqueous antifreeze 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.
A long problem here. You can help by telling me what you don't understand about it before I spent half the night working all of it out in detail.
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40% by mass means 40 g ethylene glycol + 60 g water.
Convert 40 g ethylene glycol to moles. moles = grams/molar mass.
molality = moles/kg solvent(60 g water = 0.060 kg).
For molarity, use mols/L of solution.
Look up mole fraction.
That's Xeth gly = (neth gly/total moles.
To find the molality (m), molarity (M), and mole fraction (X) of the ethylene glycol solution, we need to calculate the number of moles of ethylene glycol first.
Given:
Mass percent of ethylene glycol (C2H6O2) = 40.0%
Density of the solution = 1.05 g/cm3
Step 1: Calculate the mass of the solution.
Let's assume we have 100 g of the solution.
Mass of the solution = 100 g
Step 2: Calculate the mass of ethylene glycol.
Mass of ethylene glycol = Mass percent × Mass of the solution
Mass of ethylene glycol = 40.0% × 100 g = 40 g
Step 3: Calculate the moles of ethylene glycol.
Moles of ethylene glycol = Mass of ethylene glycol / Molar mass of ethylene glycol
The molar mass of ethylene glycol (C2H6O2) = (2 × Atomic mass of carbon) + (6 × Atomic mass of hydrogen) + (2 × Atomic mass of oxygen)
The atomic masses are:
Carbon (C) = 12.01 g/mol
Hydrogen (H) = 1.01 g/mol
Oxygen (O) = 16.00 g/mol
Molar mass of ethylene glycol = (2 × 12.01 g/mol) + (6 × 1.01 g/mol) + (2 × 16.00 g/mol)
Molar mass of ethylene glycol = 62.07 g/mol
Moles of ethylene glycol = 40 g / 62.07 g/mol
Step 4: Calculate the moles of water.
To calculate the moles of water, we subtract the moles of ethylene glycol from the total moles of the solution.
Moles of water = total moles of the solution - moles of ethylene glycol
Total moles of the solution = Mass of the solution / Molar mass of the solution
The molar mass of the solution is the sum of the molar masses of the ethylene glycol and water.
Molar mass of the solution = (Mass of ethylene glycol / Moles of ethylene glycol) + (Mass of water / Moles of water)
The molar mass of water (H2O) = (2 × Atomic mass of hydrogen) + Atomic mass of oxygen
The atomic masses are:
Hydrogen (H) = 1.01 g/mol
Oxygen (O) = 16.00 g/mol
Molar mass of water = (2 × 1.01 g/mol) + 16.00 g/mol
Molar mass of water = 18.02 g/mol
Total moles of the solution = 100 g / (40 g/mol + (100 g - 40 g) / 18.02 g/mol)
Step 5: Calculate molality (m).
Molality (m) = Moles of solute (ethylene glycol) / Mass of solvent (water) in kg
Mass of solvent (water) = Mass of the solution - Mass of solute
Mass of solvent (water) = 100 g - 40 g = 60 g
Mass of solvent (water) in kg = 60 g / 1000 g/kg
Step 6: Calculate molarity (M).
Molarity (M) = Moles of solute (ethylene glycol) / Volume of solution in liters
Since the density of the solution is given, we can calculate the volume of the solution.
Volume of solution = Mass of the solution / Density of the solution
Volume of solution = 100 g / 1.05 g/cm3
Volume of solution = 95.24 cm3 = 0.09524 L
Step 7: Calculate mole fraction (X).
Mole fraction (X) of ethylene glycol = Moles of ethylene glycol / Total moles of the solution
Now, using the calculated values, we can find the answers:
Moles of ethylene glycol = 40 g / 62.07 g/mol
Total moles of the solution = 100 g / (40 g/mol + (100 g - 40 g) / 18.02 g/mol)
Mass of solvent (water) in kg = 60 g / 1000 g/kg
Molality (m) = Moles of solute (ethylene glycol) / Mass of solvent (water) in kg
Molarity (M) = Moles of solute (ethylene glycol) / Volume of solution in liters
Mole fraction (X) of ethylene glycol = Moles of ethylene glycol / Total moles of the solution
To calculate the molality (m), molarity (M), and mole fraction (X) of ethylene glycol in the aqueous antifreeze solution, we need to use the given information about the concentration 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 molality, we need to assume a known mass of solution. Let's assume we have 100 g of the solution.
Mass of ethylene glycol = 40.0% of 100 g = 40.0 g
Mass of water = Total mass of solution - Mass of ethylene glycol = 100 g - 40.0 g = 60.0 g
The molar mass of ethylene glycol (C2H6O2) can be calculated as follows:
2 * 12.01 g/mol (C) + 6 * 1.01 g/mol (H) + 2 * 16.00 g/mol (O) = 62.07 g/mol
Number of moles of ethylene glycol = Mass of ethylene glycol / Molar mass = 40.0 g / 62.07 g/mol
Number of moles of water = Mass of water / Molar mass (of water) = 60.0 g / 18.015 g/mol
Molality (m) = Moles of ethylene glycol / Mass of water (in kg) = (40.0 g / 62.07 g/mol) / (60.0 g / 1000) kg
2. Molarity (M):
Molarity is defined as the number of moles of solute per liter of solution.
Number of moles of ethylene glycol = Mass of ethylene glycol / Molar mass = 40.0 g / 62.07 g/mol
Volume of the solution = Mass of the solution / Density of the solution = 100 g / 1.05 g/cm^3 = 95.24 cm^3 = 0.09524 L
Molarity (M) = Moles of ethylene glycol / Volume of the solution (in L) = (40.0 g / 62.07 g/mol) / 0.09524 L
3. Mole Fraction (X):
Mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the solution.
Number of moles of ethylene glycol = Mass of ethylene glycol / Molar mass = 40.0 g / 62.07 g/mol
Number of moles of water = Mass of water / Molar mass (of water) = 60.0 g / 18.015 g/mol
Total moles of all components = Moles of ethylene glycol + Moles of water
Mole fraction (X) of ethylene glycol = Moles of ethylene glycol / Total moles of all components
Now, you can substitute the values in the respective formulas to calculate the numeric values of molality, molarity, and mole fraction of ethylene glycol in the aqueous antifreeze solution.