Question

Consider two solutions, one formed by adding $150\mathrm{g}$ of glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ to $1\mathrm{L}$ of water and the other formed by adding $150\mathrm{g}$ of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ to $1\mathrm{L}$ of water. Calculate the vapor pressure for each solution at ${25}^{\circ }\mathrm{C};$ the vapor pressure of pure water at this temperature is $3.17\mathrm{kPa}$.

Short answer

The vapor pressure of the glucose solution at 25°C is approximately 3.12 kPa, and the vapor pressure of the sucrose solution at 25°C is approximately 3.15 kPa.

Step-by-step solution

Calculate the moles of solute (glucose and sucrose)

To calculate the moles of solute, we use the formula: moles = mass (g) / molar mass (g/mol) For glucose, the mass is 150 g and the molar mass is: $C_6{H}_{12}{O}_6=6(12.01)+12(1.01)+6(16.00)=180.18\mathrm{g}/\mathrm{mol}$ For sucrose, the mass is 150 g and the molar mass is: $C_{12}{H}_{22}{O}_{11}=12(12.01)+22(1.01)+11(16.00)=342.30\mathrm{g}/\mathrm{mol}$ Now, calculate the moles of both solutes: moles of glucose = 150 g / 180.18 g/mol ≈ 0.8329 mol moles of sucrose = 150 g / 342.30 g/mol ≈ 0.4382 mol

Calculate the moles of solvent (water)

To find the moles of water, we first need to convert the volume of water from liters to grams. The density of water is about 1 g/mL, so 1 L water = 1000 g water. The molar mass of water is 18.02 g/mol. moles of water = 1000 g / 18.02 g/mol ≈ 55.5 mol

Calculate the mole fraction of solvent (water) in each solution

Mole fraction is calculated as the ratio of moles of a particular component to the total moles in the solution. Calculate the mole fraction of water in both solutions using the moles of solutes and solvent calculated in steps 1 and 2: Mole fraction of water in glucose solution = moles of water / (moles of water + moles of glucose) = 55.5 / (55.5 + 0.8329) ≈ 0.9852 Mole fraction of water in sucrose solution = moles of water / (moles of water + moles of sucrose) = 55.5 / (55.5 + 0.4382) ≈ 0.9921

Calculate the vapor pressure of each solution

To find the vapor pressure (P_solution) of each solution, we will use Raoult's Law: P_solution = mole fraction of water × P_pure_water At 25°C, the vapor pressure of pure water is 3.17 kPa. For the glucose solution: P_solution_glucose = 0.9852 × 3.17 kPa ≈ 3.12 kPa For the sucrose solution: P_solution_sucrose = 0.9921 × 3.17 kPa ≈ 3.15 kPa #Conclusion# The vapor pressure of the glucose solution at 25°C is approximately 3.12 kPa, and the vapor pressure of the sucrose solution at 25°C is approximately 3.15 kPa.

Key concepts

Raoult's Law

Raoult's Law is a fundamental principle used to determine the vapor pressure of a solution. It states that the vapor pressure of a solution ($P_{\text{solution}}$) is directly proportional to the mole fraction of the solvent ($X_{\text{solvent}}$) in the solution and the vapor pressure of the pure solvent ($P_{\text{pure}}$) at the same temperature. The formula can be represented as:$$P_{\text{solution}}=X_{\text{solvent}}\times P_{\text{pure}}$$This law applies to ideal solutions, meaning that the solute and solvent molecules have similar intermolecular forces. Raoult's law is particularly useful when calculating the reduction in vapor pressure from adding a non-volatile solute, such as glucose or sucrose, to a solvent like water. Since the solute occupies space at the liquid’s surface, there are fewer water molecules to escape into the vapor phase, decreasing the overall vapor pressure.

Mole Fraction

The mole fraction is a key concept in understanding solutions' chemistry, and it represents the ratio of the number of moles of a particular component to the total number of moles in the mixture. It is dimensionless and provides a way to express the concentration of a component in a solution. For a binary solution like the one described in the exercise, comprising a solute and a solvent, the mole fraction is calculated for each component. For example:

• For water in a solution containing glucose, the mole fraction of water ($X_{\text{water}}$) is given by $\frac{\text{moles of water}}{\text{moles of water}+\text{moles of glucose}}$.
• Likewise, for a sucrose solution, $X_{\text{water}}$ is $\frac{\text{moles of water}}{\text{moles of water}+\text{moles of sucrose}}$.

This concept helps in determining how the presence of solute affects the total moles in the solution, thus influencing the vapor pressure according to Raoult's Law.

Molar Mass

Molar mass is an essential property when dealing with solutions, as it allows conversion between mass and moles for compounds. It is defined as the mass of one mole of a given substance, usually expressed in grams per mole (g/mol).In our example:- Glucose ($C_6{H}_{12}{O}_6$) has a molar mass of 180.18 g/mol.- Sucrose ($C_{12}{H}_{22}{O}_{11}$) has a molar mass of 342.30 g/mol.Molar mass is crucial when calculating the amount of substance added to a solution, as it provides the link between the measurable mass of the solute and its quantity in moles, which is used to calculate the mole fraction. This understanding forms the basis for determining the effect of solute on the solution’s vapor pressure as outlined in Raoult's Law.

Solutions Chemistry

Solutions Chemistry involves understanding the properties and behaviors of solutions, which are homogeneous mixtures of two or more substances. In our context, the focus is on how a solute (glucose or sucrose) interacts with a solvent (water) and impacts properties like vapor pressure. Key aspects of solutions chemistry include:

• Concentration: Measured in units like molarity or mole fraction, which describe how much solute is present relative to the solvent.
• Interactions: The nature of solute-solvent interactions affects solution properties. Non-volatile solutes lower a solution’s vapor pressure, as seen through Raoult's Law.
• Impact on Physical Properties: Adding solutes can impact boiling point, freezing point, and vapor pressure. This phenomenon is called colligative properties, which depend on the number of solute particles in the solution, not their specific identity.

Understanding these concepts allows chemists to predict how a solution will behave under various conditions, such as changes in temperature or pressure, thereby informing many practical applications in fields ranging from pharmaceuticals to environmental science.