Question

A binary solution at \(25^{\circ} \mathrm{C}\) consists of \(59 \mathrm{~kg}\) of ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(41 \mathrm{~kg}\) of water. The respective partial molal volumes are \(0.0573\) and \(0.0172 \mathrm{~m}^{3} / \mathrm{kmol}\). Determine the total volume, in \(\mathrm{m}^{3}\). Compare with the volume calculated using the molar specific volumes of the pure components, each a liquid at \(25^{\circ} \mathrm{C}\), in the place of the partial molal volumes.

Short answer

The total volume using partial molal volumes is 0.1127 m³. Using pure component specific volumes, the total volume is 0.1162 m³.

Step-by-step solution

- Molar Masses

Calculate the molar masses of ethyl alcohol \((C_{2}H_{5}OH)\) and water (H_{2}O). The molar mass of ethyl alcohol is obtained as follows: \[C_{2}H_{5}OH = (2 \times 12) + (6 \times 1) + 16 = 46 \ \mathrm{kg/kmol}\] The molar mass of water is obtained as follows: \[H_{2}O = (2 \times 1) + 16 = 18 \ \mathrm{kg/kmol}\]

- Moles Calculation

Determine the number of moles of each component using the formula \(n = \frac{m}{M}\), where \(m\) is the mass and \(M\) is the molar mass. For ethyl alcohol: \[n_{\text{ethyl alcohol}} = \frac{59 \ \mathrm{kg}}{46 \ \mathrm{kg/kmol}} = 1.283 \ \mathrm{kmol}\] For water: \[n_{\text{water}} = \frac{41 \ \mathrm{kg}}{18 \ \mathrm{kg/kmol}} = 2.278 \ \mathrm{kmol}\]

- Use Partial Molal Volumes

Calculate the total volume using partial molal volumes. The total volume \(V_{\text{total}}\) using partial molal volumes is given by: \[V_{\text{total}} = (n_{\text{ethyl alcohol}} \times V_{\text{ethyl alcohol}}) + (n_{\text{water}} \times V_{\text{water}})\] Substitute the values: \[V_{\text{total}} = (1.283 \ \mathrm{kmol} \times 0.0573 \ \mathrm{m^3/kmol}) + (2.278 \ \mathrm{kmol} \times 0.0172 \ \mathrm{m^3/kmol}) \ \approx 0.0735 \ \mathrm{m^3} + 0.0392 \ \mathrm{m^3} = 0.1127 \ \mathrm{m^3}\]

- Use Molar Specific Volumes for Comparison

Calculate the volumes using the molar specific volumes of the pure components. Assume pure component molar volumes: \V_{\text{ethyl alcohol}}^{pure} = 0.0585 \ \mathrm{m^3/kmol}\ and \V_{\text{water}}^{pure} = 0.0181 \ \mathrm{m^3/kmol}\. The total volume \(V_{\text{total}}^{pure}\) is then: \[V_{\text{total}}^{pure} = (n_{\text{ethyl alcohol}} \times V_{\text{ethyl alcohol}}^{pure}) + (n_{\text{water}} \times V_{\text{water}}^{pure}) \ \approx (1.283 \ \mathrm{kmol} \times 0.0585 \ \mathrm{m^3/kmol}) + (2.278 \ \mathrm{kmol} \times 0.0181 \ \mathrm{m^3/kmol}) = 0.0750 \ \mathrm{m^3} + 0.0412 \ \mathrm{m^3} = 0.1162 \ \mathrm{m^3}\]

Key concepts

Molar Mass

To understand how to calculate the volume of a binary solution, it’s first important to grasp the concept of molar mass. The molar mass of a substance is the mass of one mole of that substance. This quantity is crucial because it allows us to convert between the mass of a substance and the amount of substance in moles, which is a core requirement for further calculations.
Molar mass is calculated using the atomic masses of each element in the molecular formula. For example, ethyl alcohol (\text{C}_{2}\text{H}_{5}\text{OH}) has a molar mass calculated as follows:

• Carbon (\text{C}): 12 \text{g/mol}, and there are 2 carbons. So, \text{2} \times 12 = 24.
• Hydrogen (\text{H}): 1 \text{g/mol}, and there are 6 hydrogens (\text{5} from \text{C}_2\text{H}_5 and \text{1} from \text{OH}). So, \text{6} \times 1 = 6.
• Oxygen (\text{O}): 16 \text{g/mol}, and there is 1 oxygen. So, \text{1} \times 16 = 16.

Therefore, the molar mass of ethyl alcohol is:
\[C_{2}H_{5}OH = 24 + 6 + 16 = 46 \text{kg/kmol}\]
Similarly, the molar mass of water (\text{H}_{2}\text{O}) is:

• Hydrogen (\text{H}): 1 \text{g/mol}, and there are 2 hydrogens. So, \text{2} \times 1 = 2.
• Oxygen (\text{O}): 16 \text{g/mol}, and there is 1 oxygen. So, \text{1} \times 16 = 16.

Thus, the molar mass of water is:
\[H_{2}O = 2 + 16 = 18 \text{kg/kmol}\]
Understanding these calculations is fundamental for solving more complex problems like volume calculations.

Moles Calculation

Given the masses of each component in the solution, we can calculate the number of moles by using the formula:
\[n = \frac{m}{M}\]where \(n\) is the number of moles, \(m\) is the mass of the component, and \(M\) is the molar mass.
For ethyl alcohol with a mass of 59 kg:

• Using its molar mass:\(\frac{59 \text{kg}}{46 \text{kg/kmol}} = 1.283 \text{kmol}\)

Similarly, for water with a mass of 41 kg:

• Using its molar mass:\(\frac{41 \text{kg}}{18 \text{kg/kmol}} = 2.278 \text{kmol}\)

Calculating the moles allows us to proceed with other calculations, such as determining the total volume based on partial molal volumes or molar specific volumes. This step is essential because it bridges the mass of components with their respective volume contributions in the solution.

Partial Molal Volumes

Partial molal volumes define the contribution each component makes to the total volume of the solution. They are specific to each component and can vary with the composition of the mixture. Given:

• Partial molal volume of ethyl alcohol: 0.0573 \text{m}^{3}/\text{kmol}
• Partial molal volume of water: 0.0172 \text{m}^{3}/\text{kmol}

The total volume \(V_{\text{total}}\) of the solution can be determined using the following formula:
\[V_{\text{total}} = (n_{\text{ethyl alcohol}} \times V_{\text{ethyl alcohol}}) + (n_{\text{water}} \times V_{\text{water}})\]Replacing with the calculated values:
\[V_{\text{total}} = (1.283 \text{kmol} \times 0.0573 \text{m}^{3}/\text{kmol}) + (2.278 \text{kmol} \times 0.0172 \text{m}^{3}/\text{kmol}) = 0.0735 \text{m}^{3} + 0.0392 \text{m}^{3} = 0.1127 \text{m}^{3}\]
This calculation signifies how much volume each mole of the components occupies, contributing to the total volume of 0.1127 \text{m}^{3} for the solution.

Total Volume Calculation

To verify our result with partial molal volumes, we should calculate the total volume using molar specific volumes of pure components.
Assume the molar specific volumes for pure ethyl alcohol and water at 25°C are:

• Ethyl alcohol: 0.0585 \text{m}^{3}/\text{kmol}
• Water: 0.0181 \text{m}^{3}/\text{kmol}

Using these pure volumes, the total volume is calculated as follows:
\[V_{\text{total}}^{\text{pure}} = (n_{\text{ethyl alcohol}} \times V_{\text{ethyl alcohol}}^{\text{pure}}) + (n_{\text{water}} \times V_{\text{water}}^{\text{pure}})\]Replacing with the numbers we calculated before:
\[V_{\text{total}}^{\text{pure}} = (1.283 \text{kmol} \times 0.0585 \text{m}^{3}/\text{kmol}) + (2.278 \text{kmol} \times 0.0181 \text{m}^{3}/\text{kmol}) = 0.0750 \text{m}^{3} + 0.0412 \text{m}^{3} = 0.1162 \text{m}^{3}\]Comparing both results:

• Total volume using partial molal volumes: 0.1127 \text{m}^{3}
• Total volume using pure component volumes: 0.1162 \text{m}^{3}

The slight difference arises due to interactions between the molecules in the solution. The partial molal volumes accommodate these interactions better, making them a more precise measure for solution calculations. The calculation with pure component volumes assumes no such interactions, which is an approximation.