Question

An aqueous antifreeze solution is 40.0\(\%\) ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) by mass. The density of the solution is 1.05 \(\mathrm{g} / \mathrm{cm}^{3}\) . Calculate the molality, molarity, and mole fraction of the ethylene glycol.

Short answer

The molality of the ethylene glycol in the solution is 10.73 mol/kg, its molarity is 6.76 mol/L, and its mole fraction is 0.162.

Step-by-step solution

Find the mass of ethylene glycol and water in the solution

Let's assume we have 100 g of the solution. Since it is 40.0% ethylene glycol by mass, it means there are 40.0 g of ethylene glycol and 60.0 g of water in the solution.

Convert mass of ethylene glycol to moles

Now, we will need to determine the number of moles of ethylene glycol in the solution. To do that, we should use the molar mass of ethylene glycol: Molar mass of C₂H₆O₂ = (2 × 12.01) + (6 × 1.01) + (2 × 16.00) = 62.07 g/mol Then, we can convert the mass of ethylene glycol to moles: Moles of ethylene glycol = (40.0 g) / (62.07 g/mol) = 0.644 mol

Calculate the molality of the solution

Molality is defined as the number of moles of solute per kilogram of solvent. Since we already know the moles of ethylene glycol and mass of water, we can calculate the molality: Molality = (moles of ethylene glycol) / (mass of water in kg) Molality = (0.644 mol) / (0.060 kg) = 10.73 mol/kg

Calculate the molarity of the solution

To calculate the molarity, we need to find the number of moles of ethylene glycol per liter of solution. We are given the density of the solution (1.05 g/cm³). First, let's convert the density to g/mL and find the volume of the 100 g solution: Density = (1.05 g/cm³) × (1 cm³/mL) = 1.05 g/mL Volume of 100 g solution = (100 g) / (1.05 g/mL) = 95.24 mL = 0.09524 L Now, we can calculate the molarity: Molarity = (moles of ethylene glycol) / (volume of solution in liters) Molarity = (0.644 mol) / (0.09524 L) = 6.76 mol/L

Calculate the mole fraction of ethylene glycol

Mole fraction is defined as the ratio of the number of moles of a component to the total number of moles in the solution. To find the mole fraction of ethylene glycol, we need to find the total number of moles in the solution: Moles of water = (60.0 g) / (18.02 g/mol) = 3.329 mol Total moles = moles of ethylene glycol + moles of water Total moles = 0.644 mol + 3.329 mol = 3.973 mol Now, we can calculate the mole fraction of ethylene glycol: Mole fraction = (moles of ethylene glycol) / (total moles) Mole fraction = (0.644 mol) / (3.973 mol) = 0.162 Thus, the molality of the solution is 10.73 mol/kg, the molarity is 6.76 mol/L, and the mole fraction of ethylene glycol is 0.162.

Key concepts

Molality

Molality is a concentration unit used in chemistry to describe the amount of solute in a solution. Unlike molarity, molality is defined by the number of moles of solute per kilogram of solvent, not the entire solution. This can be especially useful when dealing with solutions where temperature changes, since volume can change with temperature, while mass does not.

To calculate molality, use the formula:
\[\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\]

• Moles of Solute: This is calculated by dividing the mass of the solute by its molar mass.
• Mass of Solvent: This needs to be converted into kilograms if not already in this unit.

In our ethylene glycol solution example, we used 40.0 g of ethylene glycol, converted it to 0.644 moles with its molar mass of 62.07 g/mol, and had 60.0 g of water (0.060 kg) as the solvent. Substituting these into the formula gives us a molality of 10.73 mol/kg.

Molarity

Molarity is another way to express the concentration of a solution, defined as the number of moles of solute per liter of solution. This is widely used in many chemical calculations and experiments.

The formula for calculating molarity is:
\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]

• Moles of Solute: Same as in molality, this is the number of moles of the substance being dissolved.
• Volume of Solution: Measured in liters, and unlike molality, includes the entire solution.

In the exercise, the density of the solution (1.05 g/mL) helps us convert the mass of the solution to volume. For 100 g of solution, we end up with a volume of 95.24 mL, or 0.09524 L. From this, we calculate the molarity as 6.76 mol/L using the 0.644 moles of ethylene glycol.

Mole Fraction

Mole fraction is a measure of concentration that expresses the ratio of the moles of one component to the total moles of all components in the solution. Unlike molality and molarity, mole fraction is a dimensionless quantity.

The formula to calculate mole fraction is:
\[\text{Mole Fraction} = \frac{\text{moles of component}}{\text{total moles in solution}}\]

• Moles of Component: The number of moles of the substance of interest.
• Total Moles: This includes both the solute and the solvent.

In the given scenario, the moles of ethylene glycol are 0.644, and for water, we have calculated 3.329 moles. With these, the total moles in the solution come to 3.973 moles. Hence, the mole fraction of ethylene glycol calculations resulted in a value of 0.162.