Problem 65 - Chemistry The Central Science

Question

(a) Calculate the vapor pressure of water above a solution prepared by adding 22.5 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 \(40^{\circ} \mathrm{C}\) .

Step-by-Step Solution

Answer

a) The vapor pressure of water above the lactose solution at 338 K is calculated using the molality of the lactose solution and Raoult's Law. We find that the molality is \(\frac{moles\,of\,lactose}{0.200\,kg}\), and then use the mole fraction of water and the vapor pressure of pure water at 338 K to find the vapor pressure: \(P_{water} = X_{water} * P_{water}^{*}\). b) To calculate the mass of propylene glycol to be added to reduce the vapor pressure by 2.88 torr at $40^{\circ} \mathrm{C}$, we first determine the molality of the propylene glycol solution, then use this to find the moles of propylene glycol in the mixture. Finally, we find the mass of propylene glycol needed by multiplying the moles by the molar mass (76 g/mol).

Step 1: a) Calculate molality of lactose solution

: First, we must calculate the molality of the lactose solution which is the moles of solute (lactose) per kilogram of solvent (water). Molality, denoted as "m", is given by: \(m = \frac{moles\,of\,solute}{mass\,of\,solvent\,(kg)}\) 1. Calculate moles of lactose (\(C_{12} H_{22} O_{11}\)): Lactose molar mass = 12(12) + 22 + 16(11) = 342 g/mol moles of lactose = mass (22.5 g) / molar mass (342 g/mol) 2. Convert mass of water to kg: mass of water = 0.200 kg 3. Calculate molality: \(m = \frac{moles\,of\,lactose}{0.200\,kg}\)

Step 2: a) Find vapor pressure of water above the solution

: After finding the molality of the lactose solution, we can use Raoult's Law to determine the vapor pressure of water above the solution at 338 K. Raoult's Law states: \(P_{A} = X_{A} * P_{A}^{*}\) where, \(P_{A}\) = partial pressure above the solution \(X_{A}\) = mole fraction of the solvent \(P_{A}^{*}\) = pure solvent vapor pressure at 338 K 1. Calculate mole fraction of water (\(X_{water}\)): \(X_{water} = \frac{moles\,of\,water}{(moles\,of\,water + moles\,of\,lactose)}\) 2. Find the vapor pressure of pure water at 338 K, which can be found in Appendix B 3. Calculate the vapor pressure above the solution: \(P_{water} = X_{water} * P_{water}^{*}\)

Step 3: b) Calculate molality of the propylene glycol solution

: This time, we have the reduced vapor pressure (2.88 torr) and we must find the mass of propylene glycol (\(C_{3}H_{8}O_{2}\)). First, find the molality of the solution required to achieve the reduced vapor pressure. Raoult's Law: \(P_{water} - 2.88\,torr = X_{propylene\,glycol} * P_{water}\) Solve for the molality of propylene glycol: \(m_{propylene\,glycol} = \frac{moles\,of\,propylene\,glycol}{mass\,of\,solvent\,(kg)}\)

Step 4: b) Calculate the mass of propylene glycol to be added

: Once we have determined the molality of the propylene glycol solution, we can find the mass of propylene glycol needed to achieve this molality. 1. Calculate moles of propylene glycol using molality and mass of solvent (0.340 kg): \(moles\,of\,propylene\,glycol = m_{propylene\,glycol} * mass\,of\,solvent\,(0.340\,kg)\) 2. Find mass of propylene glycol using moles and its molar mass (\(C_{3}H_{8}O_{2}\): Propylene glycol molar mass = 3(12) + 8 + 16(2) = 76 g/mol mass of propylene glycol = moles of propylene glycol × molar mass (76 g/mol)

Key Concepts

Raoult's Law

Raoult's Law serves as a fundamental principle in understanding vapor pressures of solutions. It states that the partial vapor pressure of a component in an ideal solution is directly proportional to its mole fraction. This can be expressed as: \( P_{A} = X_{A} \times P_{A}^{*} \)
where:

• \( P_{A} \) represents the partial pressure of the solvent above the solution.
• \( X_{A} \) stands for the mole fraction of the solvent.
• \( P_{A}^{*} \) is the vapor pressure of the pure solvent at the same temperature.

Raoult's Law applies well to ideal solutions where the interactions between all particles are similar. However, real-world solutions might deviate due to solute-solvent interactions.
Understanding Raoult’s Law helps to predict how much the vapor pressure of a solvent decreases when a non-volatile solute is dissolved in it. Solutions that follow Raoult's Law are characterized as having little to no volume change or heat effect upon mixing.

Molality

Molality is a concentration term used to describe a solution's solute concentration in relation to the solvent's mass. It is not dependent on temperature, which makes it particularly useful for colligative properties calculations.
Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent, expressed as:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
In the exercise, for instance, we calculated the molality of a lactose solution by determining the moles of lactose and dividing it by the mass of water used as the solvent in kilograms.

• Benefits of Using Molality:
  • Independent of temperature changes.
  • Useful for comparing different solutions under various conditions.
  • Integral in calculating colligative properties like boiling point elevation and freezing point depression.

Because molality considers the mass of the solvent rather than the solution’s volume, it offers consistent measurements across temperature fluctuations.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture, defined as the ratio of the number of moles of one component to the total number of moles in the mixture. It is unitless and always less than or equal to one.
For a solvent in a solution, the formula for mole fraction (\( X_A \)) is:\[ X_{A} = \frac{\text{moles of solvent}}{\text{total moles of all components}} \]
This concept was used in the exercise to calculate the mole fraction of water in the lactose solution. Knowing the mole fraction is critical for applying Raoult’s Law and estimating the lowering of vapor pressure in solutions where a non-volatile solute is present.

• Advantages of assessing via Mole Fraction:
  • Dimensionless quantity, making it straightforward to use in calculations.
  • Unaffected by temperature and pressure variations.

Mole fraction provides a universal measure that can easily be converted to other concentration terms and is vital in theoretical and practical chemistry applications.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the number of solute particles, rather than their identity. These include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
In the provided exercise, we explored the colligative property of vapor pressure lowering using Raoult’s Law. By calculating how the vapor pressure of water decreases upon adding lactose, it highlights the principle that adding a non-volatile solute to a solvent reduces its vapor pressure.

• Key Concepts in Colligative Properties:
  • Relate to the quantity of solute molecules in a solution.
  • Help explain phenomena like antifreeze applications and cooking in higher altitudes.
  • Utilized in laboratory conditions to determine molar masses.

Although the solute's chemical nature is ignored in calculations, the number of solute particles remains central to understanding colligative properties. This explains why ionic compounds exhibit a more significant impact due to their dissociation into more particles.