Question

(a) What is an ideal solution? (b) The vapor pressure of pure water at \(60^{\circ} \mathrm{C}\) is 149 torr. The vapor pressure of water over a solution at \(60^{\circ} \mathrm{C}\) containing equal numbers of moles of water and ethylene glycol (a nonvolatile solute) is 67 torr. Is the solution ideal according to Raoult's law? Explain.

Short answer

(a) An ideal solution is a solution in which the interaction between the solute and solvent molecules is the same as that between the solute-solute and solvent-solvent molecules. In an ideal solution, the vapor pressure of the solvent over the solution is proportional to its mole fraction, following Raoult's law. (b) We calculate the mole fraction of water (\(x_{H_2O}\)) and ethylene glycol (\(x_{EG}\)) as 0.5 for both. Using Raoult's law, the theoretical vapor pressure of water in the solution is 74.5 torr, while the given vapor pressure is 67 torr. Since these values are not equal, the solution does not follow Raoult's law, and thus, it is not an ideal solution.

Step-by-step solution

(a) Definition of an Ideal Solution

An ideal solution is a solution in which the interaction between the solute and solvent molecules is the same as that between the solute-solute and solvent-solvent molecules. In an ideal solution, the vapor pressure of the solvent over the solution is proportional to its mole fraction in the solution, following Raoult's law.

(b) Applying Raoult's Law

Step 1: Calculate the mole fraction of water and ethylene glycol The problem states that there are equal numbers of moles of water and ethylene glycol in the solution. Let's denote the mole fraction of water as \(x_{H_2O}\) and the mole fraction of ethylene glycol as \(x_{EG}\). Since there are equal moles of both components: \[x_{H_2O} = x_{EG}\] The sum of mole fractions in any solution is always equal to 1: \[x_{H_2O} + x_{EG} = 1\] Since \(x_{H_2O} = x_{EG}\), we can deduce: \[2x_{H_2O} = 1\] \[x_{H_2O} = x_{EG} = 0.5\] Step 2: Apply Raoult's Law and calculate the theoretical vapor pressure We are given the vapor pressure of pure water at \(60^{\circ}\mathrm{C}\) as 149 torr. Using Raoult's law, we can calculate the vapor pressure of water in the solution (\(P_{H_2O}\)): \[P_{H_2O} = x_{H_2O} \cdot P^{\circ}_{H_2O}\] Where \(P^{\circ}_{H_2O}\) is the vapor pressure of pure water at the given temperature. \[P_{H_2O} = 0.5 * 149 \, \mathrm{torr}\] \[P_{H_2O} = 74.5 \, \mathrm{torr}\] Step 3: Compare the theoretical vapor pressure with the given vapor pressure We calculated the theoretical vapor pressure of water in the solution as 74.5 torr, using Raoult's law, while the given vapor pressure of water is 67 torr. Since these values are not equal, the solution does not follow Raoult's law, and thus, it is not an ideal solution.