Question

What volume of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right),\) a nonelectrolyte,must be added to 15.0 \(\mathrm{L}\) water to produce an antifreeze solution with a freezing point of \(-25.0^{\circ} \mathrm{C} ?\) What is the boiling point of this solution? (The density of ethylene glycol is \(1.11 \mathrm{g} / \mathrm{cm}^{3},\) and the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3} . )\)

Short answer

To find the volume of ethylene glycol needed to produce an antifreeze solution with a freezing point of -25.0°C, we need to first find the molality, then convert it to moles and finally to volume. The molality is calculated as: \(m = \frac{25.0}{1.86}\). This gives us the moles of ethylene glycol needed: \(Moles_{ethylene\ glycol} = \frac{25.0}{1.86} \times \frac{15000}{1000} \). Then, we can find the mass of ethylene glycol and its volume: \(Volume_{ethylene\ glycol} = \frac{Mass_{ethylene\ glycol}}{Density_{ethylene\ glycol}}\). To find the boiling point of this solution, we can use the boiling point elevation formula: \(\Delta T_b = 1 \cdot 0.512 \cdot \frac{25.0}{1.86}\). The final boiling point would be the sum of the boiling point elevation and the normal boiling point of water (100°C): \(Boiling\ point_{solution} = 100 + \Delta T_b\).

Step-by-step solution

Find the molality needed to reach the desired freezing point

First, we'll determine the molality (moles of solute per kilogram of solvent) needed for the solution to have a freezing point of -25.0°C. Freezing point depression can be calculated using the formula: \(-\Delta T_f = i \cdot K_f \cdot m\) Where \(\Delta T_f\) represents the change in freezing point (in this case, 25.0°C, since the normal freezing point of water is 0°C), \(i\) is the van't Hoff factor (number of particles into which a solute dissociates; since ethylene glycol is a nonelectrolyte, its van't Hoff factor is 1), \(K_f\) is the molal freezing-point depression constant for water (1.86 °C·kg/mol), and \(m\) is the molality. To find the molality, we can rearrange the formula and solve: \(m = \frac{-\Delta T_f}{i \cdot K_f} = \frac{25.0}{1 \cdot 1.86}\)

Calculate the moles of ethylene glycol

Now that we have the molality, we can find the number of moles of ethylene glycol needed. We know that there are 15.0 L of water in the solution, which has a density of 1.00 g/cm³. First, we need to convert the volume of water into mass: \(Mass_{water} = 15.0 \ \mathrm{L} \times 1000 \ \mathrm{cm}^{3}/\mathrm{L} \times 1.00 \ \mathrm{g}/\mathrm{cm}^{3} = 15000 \ g\) Now we can use the molality to calculate the moles of ethylene glycol: \(Moles_{ethylene\ glycol} = molality \times Mass_{solvent} = \frac{25.0}{1.86} \times \frac{15000}{1000} \)

Convert moles of ethylene glycol to volume

Now, we'll convert the moles of ethylene glycol into volume. First, we need to determine the gram-molecular mass of ethylene glycol: Molar mass of ethylene glycol = (2 × 12.01) + (6 × 1.01) + (2 × 16.00) = 62.08 g/mol We can now find the mass of ethylene glycol needed: \(Mass_{ethylene\ glycol} = Moles_{ethylene\ glycol} \times Molar\ mass_{ethylene\ glycol}\) And finally, we can determine the volume of ethylene glycol using its given density: \(Volume_{ethylene\ glycol} = \frac{Mass_{ethylene\ glycol}}{Density_{ethylene\ glycol}}\)

Determine the boiling point of the solution

Now that we've found the volume of ethylene glycol needed, we can determine the boiling point of this solution using the boiling point elevation formula: \(\Delta T_b = i \cdot K_b \cdot m \) Where \(K_b\) is the molal boiling-point elevation constant for water (0.512 °C·kg/mol). Since we've already calculated the molality in Step 1, we can directly plug it into the formula: \(\Delta T_b = 1 \cdot 0.512 \cdot \frac{25.0}{1.86}\) The final boiling point would be the sum of the boiling point elevation and the normal boiling point of water (100°C): \(Boiling\ point_{solution} = 100 + \Delta T_b\)

Key concepts

Molality

Molality is an important concept in chemistry, especially when dealing with solutions. It is defined as the number of moles of solute per kilogram of solvent. This concentration measure helps to calculate changes in physical properties such as freezing and boiling points.

The formula to calculate molality is \[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \] This makes molality a useful concentration unit when the temperature changes, as it doesn't depend on volume, which can vary with temperature.

In our problem, we calculated the molality needed to lower the freezing point of water using the equation for freezing point depression. For a solution to freeze at \(-25.0^{\circ} \mathrm{C}\), we rearrange the formula to solve for molality. Understanding molality is crucial for accurate calculations in experimental and theoretical chemistry.

Boiling Point Elevation

Boiling point elevation is a colligative property of solutions. It occurs when a solute is added to a solvent, raising the boiling point of the resulting solution above that of the pure solvent. This change happens due to the solute particles disrupting the liquid phase, requiring more energy (heat) for the liquid to transition to gas.

The formula for calculating boiling point elevation is \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the change in boiling point, \( i \) is the van't Hoff factor, \( K_b \) is the boiling point elevation constant (specific to each solvent), and \( m \) is the molality of the solution.

In our exercise, we calculate \( \Delta T_b \) for an ethylene glycol solution. The normal boiling point of water is \( 100^{\circ} \mathrm{C} \), and the calculated elevation adds to this temperature, giving the new boiling point of the solution.

Solution Chemistry

Solution chemistry revolves around understanding how solutes (like ethylene glycol) interact with solvents (such as water) to form a solution. In this context, solutions display unique properties such as viscosity, surface tension, and boiling/freezing point alterations.

These alterations are categorized as colligative properties, depending only on the number of solute particles, not their identity. As discussed, freezing point depression and boiling point elevation are key colligative properties.

For this exercise, the solution chemistry concepts help us predict how adding ethylene glycol impacts water's freezing and boiling points. By understanding these interactions, we can determine practical applications like designing antifreeze solutions for vehicles, ensuring that engine fluids remain functional in cold climates.

Thus, mastering solution chemistry is vital in both academic and industrial applications, permitting the prediction and control of the physical properties of solutions.