Question

(a) Calculate the vapor pressure of water above a solution prepared by adding \(22.5 \mathrm{~g}\) of lactose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to \(200.0 \mathrm{~g}\) of water at \(338 \mathrm{~K}\). (Vapor-pressure data for water are given in Appendix B.) (b) Calculate the mass of propylene glycol \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\right)\) that must be added to \(0.340 \mathrm{~kg}\) of water to reduce the vapor pressure by 2.88 torr at \(40^{\circ} \mathrm{C}\).

Short answer

The vapor pressure of water above the solution prepared by adding \(22.5 \mathrm{~g}\) of lactose to \(200.0 \mathrm{~g}\) of water at \(338 \mathrm{~K}\) is approximately \(185.5\,torr\). To reduce the vapor pressure of water by \(2.88\,torr\) at \(40^{\circ} \mathrm{C}\), approximately \(74.6\,g\) of propylene glycol must be added to \(0.340 \mathrm{~kg}\) of water.

Step-by-step solution

a) Calculate the vapor pressure of water above the solution

First, we need to calculate the mole fraction of water in the solution. To do this, we need to determine the moles of lactose and water in the solution. 1. Get the molar mass of lactose: \(\mathrm{C}_{12} \mathrm{H}_{22}\mathrm{O}_{11} = 12(12.01) + 22(1.01) + 11(16.00) = 342.3 \frac{g}{mol}\). 2. Calculate the moles of lactose in the solution: \(n_{lactose} = \frac{22.5\,g}{342.3\,\frac{g}{mol}} = 0.0657\,mol\). 3. Get the molar mass of water: \(\mathrm{H}_{2} \mathrm{O}= 2(1.01) + 16.00 = 18.02 \frac{g}{mol}\). 4. Calculate the moles of water in the solution: \(n_{water} = \frac{200.0\,g}{18.02\,\frac{g}{mol}} = 11.10\, mol\). 5. Determine the mole fraction of water in the solution: \(x_{water} = \frac{n_{water}}{n_{water} + n_{lactose}} = \frac{11.10\,mol}{11.10\,mol + 0.0657\,mol} = 0.994\). 6. Find the vapor pressure of pure water at the given temperature (338 K) from Appendix B (or using a reference): \(P_{water}^* = 187\,torr\). 7. Apply Raoult's law to find the vapor pressure of water above the solution: \(P_{solution} = x_{water}P_{water}^* = (0.994)(187\,torr) = 185.5\,torr\). The vapor pressure of water above the solution is approximately 185.5 torr.

b) Calculate the mass of propylene glycol to reduce the vapor pressure by 2.88 torr

We need to find the mass of propylene glycol that, when added to \(0.340 \mathrm{~kg}\) of water, will reduce the vapor pressure of water by \(2.88\,torr\). 1. Calculate the moles of water in \(0.340 \mathrm{~kg}\) of water: \(n_{water} = \frac{340\,g}{18.02\,\frac{g}{mol}} = 18.87\, mol\). 2. Find the vapor pressure of pure water at \(40^{\circ} \mathrm{C}\) (or \(313.15\,K\)) from Appendix B (or using a reference): \(P_{water}^* = 55.28\,torr\). 3. Calculate the desired vapor pressure of the solution: \(P_{solution} = P_{water}^* - 2.88\,torr = 55.28\,torr - 2.88\,torr = 52.40\,torr\). 4. Use Raoult's law to find the mole fraction of water in the solution: \(x_{water} = \frac{P_{solution}}{P_{water}^*} = \frac{52.40\,torr}{55.28\,torr} = 0.9480\). 5. Calculate the mole fraction of propylene glycol in the solution: \(x_{propylene\,glycol} = 1-x_{water} = 0.0520\). 6. Use the relationship between the mole fractions and the number of moles to find the moles of propylene glycol needed: \[\frac{n_{propylene\,glycol}}{n_{water}} = \frac{x_{propylene\,glycol}}{x_{water}} \Rightarrow n_{propylene\,glycol} = \frac{x_{propylene\,glycol}}{x_{water}} \times n_{water} = (0.0520)(18.87\,mol) = 0.981\,mol.\] 7. Get the molar mass of propylene glycol: \(\mathrm{C}_{3} \mathrm{H}_{8}\mathrm{O}_{2} = 3(12.01) + 8(1.01) + 2(16.00) = 76.10 \frac{g}{mol}\). 8. Calculate the mass of propylene glycol needed: \(m_{propylene\,glycol} = n_{propylene\,glycol}\times M_{propylene\,glycol} = (0.981\,mol)(76.10\,\frac{g}{mol}) = 74.6\,g\). The mass of propylene glycol that needs to be added to the water to reduce the vapor pressure by \(2.88\,torr\) at \(40^{\circ} \mathrm{C}\) is approximately \(74.6\,g\).

Key concepts

Raoult's Law

Raoult's Law is pivotal in understanding how the presence of a non-volatile solute can affect the vapor pressure of a solvent. In its essence, Raoult's Law states that the partial vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in the solution. The formula is given by \( P_{solution} = x_{solvent} P_{solvent}^* \), where \( P_{solution} \) is the vapor pressure of the solvent in the solution, \( x_{solvent} \) is the mole fraction of the solvent, and \( P_{solvent}^* \) is the vapor pressure of the pure solvent.

For a solution containing a non-volatile solute, the vapor pressure will be lower than that of the pure solvent. This decrease in vapor pressure is proportional to the number of solute particles in the solution. Therefore, adding a solute to a solvent decreases the solvent's vapor pressure, a fact used in the step by step solution to find the vapor pressure of water above a lactose solution.

Mole Fraction

Understanding mole fraction is vital when it comes to calculations involving Raoult's Law and solutions in general. The mole fraction is defined as the ratio of the number of moles of a particular component to the total number of moles of all components in the solution. Mathematically, it's expressed as \( x_i = n_i / \sum{n} \), where \( x_i \) is the mole fraction of component \( i \) and \( n_i \) its number of moles.

The mole fraction is a dimensionless quantity and plays a crucial role in calculating the resulting vapor pressure in a solution. It also avoids the complexity of units since it does not rely on the mass or volume of the components, a point highlighted in the exercise when calculatinating the mole fraction of water in a solution with lactose.

Colligative Properties

Colligative properties are properties of solutions that depend on the ratio of solute particles to solvent molecules in a solution, and not on the nature of the chemical species present. These properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

The exercise requires the understanding of one of these colligative properties—vapor pressure lowering—since the addition of a solute like lactose results in a lower vapor pressure compared to the pure solvent. Colligative properties are rooted in the fact that solute particles interfere with the molecules of the solvent escaping into the vapor phase, thus affecting the overall physical behavior of the solution.

Solution Chemistry

Solution chemistry is the study of solutes dissolved in solvents to form solutions. The nature of the solute-solvent interaction and the proportion of solute to solvent play critical roles in the properties of the solution, such as conductivity, reactivity, and colligative properties.

In the provided example, lactose—a non-electrolyte—is dissolved in water. The solute (lactose) does not ionize in the solution, hence the physical properties such as vapor pressure are affected only by the number of lactose particles relative to the number of water molecules, a principle applied in both parts of the exercise to calculate the changes in vapor pressure. Such principles are essential for both theoretical calculations and practical applications, such as antifreeze in car engines or preserving food by lowering the freezing point.