Question

A solution is made by mixing \(43.9 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{C}_{6}\) in \(100.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). If the vapour pressure of pure water is 26.5 torr at a particular temperature, what is the vapour pressure of the solution at the same temperature?

Short answer

The vapour pressure of the solution is approximately 25.4 torr.

Step-by-step solution

Calculate Moles of Solute

Determine the number of moles of the solute, \( \mathrm{C}_6\mathrm{H}_{12}\mathrm{C}_6 \). The molar mass of \( \mathrm{C}_6\mathrm{H}_{12}\mathrm{C}_6 \) is 180.18 g/mol. Use the formula:\[\text{moles} = \frac{\text{mass of solute}}{\text{molar mass of solute}}\]Substitute the given values:\[\text{moles of } \mathrm{C}_6\mathrm{H}_{12}\mathrm{C}_6 = \frac{43.9 \text{ g}}{180.18 \text{ g/mol}} \approx 0.244 \text{ moles}\]

Calculate Moles of Solvent

Determine the number of moles of the solvent, \( \mathrm{H}_2\mathrm{O} \). The molar mass of water is 18.015 g/mol. Use the formula:\[\text{moles} = \frac{100.0 \text{ g}}{18.015 \text{ g/mol}}\]Calculate to find:\[\text{moles of } \mathrm{H}_2\mathrm{O} \approx 5.55 \text{ moles}\]

Calculate Mole Fraction of Solvent

The mole fraction \( X \) of the solvent (\( \mathrm{H}_2\mathrm{O} \)) can be calculated using:\[X_{\text{H}_2\text{O}} = \frac{\text{moles of } \mathrm{H}_2\mathrm{O}}{\text{moles of } \mathrm{H}_2\mathrm{O} + \text{moles of } \mathrm{C}_6\mathrm{H}_{12}\mathrm{C}_6}\]Substitute the moles calculated:\[X_{\text{H}_2\text{O}} = \frac{5.55}{5.55 + 0.244} \approx 0.958\]

Calculate Vapour Pressure of the Solution

Use Raoult’s Law for the vapour pressure of the solution:\[P_{\text{solution}} = X_{\text{H}_2\text{O}} \times P^0_{\text{H}_2\text{O}}\]where \( P^0_{\text{H}_2\text{O}} \) is the vapour pressure of pure water (26.5 torr). Substitute the given values:\[P_{\text{solution}} = 0.958 \times 26.5 \approx 25.4 \text{ torr}\]

Key concepts

Mole Fraction

When we talk about the mole fraction in chemistry, we are discussing the ratio of moles of one component to the total moles in the solution. Imagine you're baking a cake; the mole fraction is akin to the proportion of sugar to the whole batch. In our problem, we determined the mole fraction of water, which is the solvent in this solution. Here's why the mole fraction is crucial:

• It helps calculate properties of solutions, like vapor pressure.
• It is a dimensionless number, meaning it doesn't have units, just a simple ratio.
• Its sum for all components in a mixture equals 1.

To find the mole fraction of the solvent (\(\text{H}_2\text{O}\)), we use its formula: \[\text{Mole Fraction of } \text{H}_2\text{O} = \frac{\text{moles of } \text{H}_2\text{O}}{\text{moles of } \text{H}_2\text{O} + \text{moles of } \text{C}_6\mathrm{H}_{12}\mathrm{C}_6}\]. This calculation helps us understand how much of the solution's composition is water, setting the groundwork for calculating other properties, such as vapor pressure.

Vapor Pressure Lowering

Vapor pressure lowering is an intriguing concept in the world of solution chemistry. It describes how the presence of a solute reduces the vapor pressure of the solvent. Here's what you need to know:

• Raoult's Law governs the relationship between the vapor pressure of a solution and the mole fraction of the solvent.
• Adding a non-volatile solute, like in our exercise, decreases the vapor pressure of the solvent.
• The reduction in vapor pressure is proportional to the amount of solute added.

The formula derived from Raoult's Law, \[P_\text{{solution}} = X_\text{{solvent}} \times P^0_\text{{solvent}}\], shows how the mole fraction affects the vapor pressure. In our example, the vapor pressure of pure water, \(P^0_{\text{H}_2\text{O}}\), was 26.5 torr, and the calculated vapor pressure of the solution was approximately 25.4 torr, showing the lowering effect of adding the solute.

Solution Chemistry

Solution chemistry is foundational to understanding how substances combine and interact on a molecular level. Here are some key points about solutions:

• Solutions consist of a solute (like \(\text{C}_6\text{H}_{12}\text{C}_6\)) dissolved in a solvent (such as water).
• They are homogeneous mixtures, meaning their composition is uniform throughout.
• The formation of a solution involves interactions between solute and solvent molecules.

In our case, we examined how a sugar-like solute combines with water to drive changes in properties like vapor pressure. This specific blend showcases the fact that adding solute affects the solvent's physical properties through phenomena like vapor pressure lowering, an essential aspect of Raoult's Law, and underscores the importance of mole fraction calculations in solution chemistry.

Calculating Moles

Calculating moles is a foundational skill in chemistry, ensuring accurate measurements in chemical reactions and solutions. To calculate moles:

• Use the formula: \(\text{moles} = \frac{\text{mass of substance}}{\text{molar mass}}\).
• This method lets you convert grams (a mass measurement) into moles (a particle count), essential for stoichiometry.
• Understanding this concept bridges the gap between the macroscopic scale (grams) and the atomic scale (particles).

Take our problem, where we computed the moles of our solute \(\text{C}_6\mathrm{H}_{12}\mathrm{C}_6\) using its mass and molar mass. Thus, we found about 0.244 moles. Similarly, for water, we calculated about 5.55 moles. Accurate mole calculations ensure precise measurements and critical in determining properties such as mole fractions and the resulting effects on solution behavior.