Question

At a certain temperature, the vapor pressure of pure benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 0.930 \(\mathrm{atm} .\) A solution was prepared by dissolving 10.0 \(\mathrm{g}\) of a nondissociating, nonvolatile solute in 78.11 \(\mathrm{g}\) of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 \(\mathrm{atm}\) . Assuming the solution behaves ideally, determine the molar mass of the solute.

Short answer

The molar mass of the solute is approximately 299 g/mol.

Step-by-step solution

Raoult's law for the solution

According to Raoult's law, the vapor pressure of the solution (P_solution) can be written as the product of the mole fraction of the solvent (X_benzene) and the vapor pressure of the pure solvent (P_pure_benzene): \[P_{solution} = X_{benzene} \cdot P_{pure\_benzene}\]

Calculate the mole fraction of benzene in the solution

We are given the vapor pressure of the pure benzene (P_pure_benzene) and the vapor pressure of the solution (P_solution). We can rearrange Raoult's law to find the mole fraction of benzene (X_benzene): \[X_{benzene} = \frac{P_{solution}}{P_{pure\_benzene}}\] Now, plug in the given values to find X_benzene: \[X_{benzene} = \frac{0.900 atm}{0.930 atm} = 0.9677\]

Calculate the mole fraction of the solute (X_solute)

Since there are only two components in the solution (benzene and the solute), their mole fractions must add up to 1: \[X_{solute} = 1 - X_{benzene}\] Using the value of X_benzene calculated in step 2: \[X_{solute} = 1 - 0.9677 = 0.0323\]

Calculate the moles of benzene and the solute in the solution

We are given the mass of benzene (78.11 g) and can look up its molar mass (78.114 g/mol): moles of benzene = mass_benzene / molar_mass_benzene moles of benzene = 78.11 g / 78.114 g/mol = 1 mol Next, we can find the moles of the solute: moles of solute = moles of benzene * (X_solute / X_benzene) moles of solute = 1 mol * (0.0323 / 0.9677) = 0.0334 mol

Calculate the molar mass of the solute

We are given the mass of the solute (10.0 g). Using the moles of the solute calculated in step 4, find the molar mass of the solute: molar_mass_solute = mass_solute / moles of solute molar_mass_solute = 10.0 g / 0.0334 mol = 299 g/mol The molar mass of the solute is approximately 299 g/mol.

Key concepts

Vapor Pressure

Vapor pressure is a fundamental concept in understanding how substances evaporate and interact in solutions. It represents the pressure exerted by the vapor of a liquid when it is in equilibrium with its liquid phase. For instance, if you have benzene in a closed container, some of the liquid benzene will evaporate and exert a pressure against the walls of the container. This is the vapor pressure.
Vapor pressure depends on temperature: as the temperature increases, the vapor pressure increases because more molecules gain enough energy to escape into the vapor phase.

• Pure vs. Solution: The vapor pressure of a pure solvent is different from that of a solution. Adding a solute reduces the number of solvent molecules at the surface, thereby lowering the vapor pressure of the solvent in the solution.
• Measurement: In our exercise, pure benzene has a vapor pressure of 0.930 atm, while the benzene solution has a slightly lower vapor pressure of 0.900 atm.

Understanding vapor pressure is crucial for predicting how a solution will behave, especially in chemical processes that involve phase changes.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture. It's the ratio of the number of moles of a specific component to the total number of moles in the solution. This is a dimensionless number and is crucial in calculations involving Raoult’s Law.
In the case of our exercise, we calculated the mole fraction of benzene (X_{benzene}) and the solute (X_{solute}) in the solution. The formula used was:\[ \text{Mole Fraction} = \frac{\text{moles of component}}{\text{total moles in solution}} \]

• For Benzene: Plugging in the pressures 0.900 atm and 0.930 atm, we calculated X_{benzene} to be approximately 0.9677.
• For Solute: Since there are only two components, the mole fractions add up to 1, allowing us to calculate X_{solute} as 0.0323.

These mole fractions provide insight into how the components interact within the solution.

Ideal Solutions

An ideal solution is a concept where the solution behaves according to Raoult’s Law without any deviation. This means the interactions between the different molecules in the solution (solvent and solute) mimic those in the pure state.
The assumption of ideality simplifies the analysis of a solution's properties because it indicates no significant changes in volume or enthalpy when mixing occurs.
In this exercise:

• The solution is assumed to be ideal, meaning benzene and the solute do not react chemically and their interactions are just like those in pure benzene.
• This assumption helps in applying Raoult’s Law accurately without extra considerations for deviations due to molecular interactions.

While real-world solutions often deviate from ideal behavior, ideal solutions provide a useful framework for understanding and predicting solution properties in introductory chemistry.

Molar Mass Calculation

Calculating the molar mass of an unknown solute is a critical application of the concepts discussed. Molar mass gives us the mass of one mole of a substance, and it's essential for determining the identity or quantity of a solute in solution.
In our example, we were given 10.0 g of solute in the solution and used the mole fraction of both the solute and benzene to find the moles of the solute:

• Step 1: From the mole fraction (X_{solute}) and the total moles, we found the moles of solute: 0.0334 mol.
• Step 2: We then calculated the molar mass using:\[ \text{Molar Mass} = \frac{\text{Mass of Solute}}{\text{Moles of Solute}} \]
• Step 3: Substituting in, the molar mass turned out to be approximately 299 g/mol.

This calculation connects theoretical concepts to practical application, allowing chemists to identify compounds or confirm the composition of a solution.