Question

At a certain temperature, the vapor pressure of pure benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 0.930 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 atm. Assuming the solution behaves ideally, determine the molar mass of the solute.

Short answer

The molar mass of the solute in the ideal solution is approximately 302 g/mol.

Step-by-step solution

Calculate the mole fraction of solute and solvent in the solution

To use Raoult's Law, we need to determine the mole fractions of the solute and solvent in the solution. Since we have the mass of the solvent (benzene) and its molar mass, we can calculate its moles. We'll denote the moles of benzene as \(n_1\) and the moles of the solute as \(n_2\), and the molar mass of the solute as \(M_2\). \[n_1 =\frac{mass_{solvent}}{molar\_mass_{solvent}}=\frac{78.11 \,\mathrm{g}}{78.11\, \mathrm{g/mol}} =1\, \mathrm{mol} \] To find the mole fraction of solute and solvent, we need to use the equation: \[ \chi_1 = \frac{n_1}{n_1+n_2} \] and \[ \chi_2 = \frac{n_2}{n_1+n_2} \] Since the sum of mole fractions is always equal to 1, we also have: \[ 1 = \chi_1 + \chi_2 \]

Apply Raoult's Law to the solution

Raoult's law states that the partial pressure of a component in an ideal solution is equal to its mole fraction multiplied by its vapor pressure in the pure state. For benzene, we can write: \[ P_1 = \chi_1 \times P^*_1 \] Where \(P_1\) is the partial pressure of benzene in the solution, \(\chi_1\) is its mole fraction, and \(P^*_1\) is its vapor pressure as a pure liquid (0.930 atm). For the solute, since it is nonvolatile, its partial pressure will be zero, so the total pressure of the solution is solely due to benzene. Therefore, we can write: \[ P_\text{solution} = P_1 = \chi_1 \times P^*_1 \] We know \(P_\text{solution}\) is 0.900 atm, so we can write: \[ 0.900\, \mathrm{atm} = \chi_1 \times 0.930\, \mathrm{atm} \]

Calculate the mole fraction of benzene

Now, we will solve for \(\chi_1\): \[ \chi_1 = \frac{0.900\, \mathrm{atm}}{0.930\, \mathrm{atm}} = 0.968 \] Since we know that \(\chi_1 + \chi_2 = 1\), we can calculate the mole fraction of the solute (\(\chi_2\)): \[ \chi_2 = 1 - \chi_1 = 1 - 0.968 = 0.032 \]

Calculate the moles and molar mass of the solute

Now, we can find the moles of the solute: \[ n_2 = \frac{\chi_2}{\chi_1} \times n_1 = \frac{0.032}{0.968} \times 1\, \mathrm{mol} = 0.0330\, \mathrm{mol} \] Now that we have the moles of the solute, we can calculate the molar mass as follows: \[ M_2 = \frac{mass_{solute}}{n_2} = \frac{10.0\, \mathrm{g}}{0.0330\, \mathrm{mol}} = 302\, \mathrm{g/mol} \] So, the molar mass of the solute is estimated to be 302 g/mol.

Key concepts

Mole Fraction

The concept of mole fraction is crucial in understanding solutions. It essentially tells us how much of a particular component is present in the mixture relative to the total amount. For any component, the mole fraction \(\chi\) is calculated as the number of moles of that component divided by the total moles of all components in the solution. To compute the mole fraction of a solute (\(\chi_2\)) or a solvent (\(\chi_1\)), we use the formulas:

• \(\chi_1 = \frac{n_1}{n_1+n_2}\)
• \(\chi_2 = \frac{n_2}{n_1+n_2}\)

This sum might always add up to 1, i.e., \(\chi_1 + \chi_2 = 1\). The mole fraction helps us apply Raoult's Law efficiently to determine partial pressures and other thermodynamic properties of solutions.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. When a nonvolatile solute is added to a solvent, the overall vapor pressure of the solution decreases. Raoult's Law quantifies this phenomenon. According to Raoult's Law, the partial vapor pressure of a component in an ideal solution is equal to its mole fraction multiplied by its vapor pressure when it is pure.This is crucial because, in our case with benzene:\(P_1 = \chi_1 \times P^*_1\)Where \(P_1\) is the partial vapor pressure of the solvent in the solution, \(\chi_1\) is the mole fraction of the solvent, and \(P^*_1\) is the vapor pressure of the pure solvent. This relation helps us understand how molecules interact in the solution and predict changes in vapor pressure when conditions change.

Molar Mass Calculation

Determining the molar mass of a solute in a solution can be performed using the mole fraction and mass relationships. Once the number of moles of solute (\(n_2\)) is known, the molar mass, denoted as \(M_2\), can be calculated using:\[ M_2 = \frac{\text{mass of solute}}{n_2} \]In our example, once the mole fraction \(\chi_2\) is determined, we can find the moles of solute via:\(n_2 = \frac{\chi_2}{\chi_1} \times n_1\)This allows us to use the total given mass of the solute to find its molar mass, which can provide insight into the identity of the solute by comparing it with known molar masses of different compounds.

Ideal Solution

An ideal solution is a special case where the properties of the solution, such as vapor pressure, can be predicted from the properties of its constituents, assuming no significant interactions other than simple mixing occur. In an ideal solution scenario, interactions between different molecules (solute-solvent) are similar to those between like molecules (solute-solute or solvent-solvent). This means that the solution exhibits properties very close to the weighted average of its components. Raoult’s Law is applicable in this context, helping predict how the addition of a solute will affect the vapor pressure of a solvent. It assumes the solution behaves ideally, meaning non-ideal behaviors like strong solute-solute attractions or solvent-solvent repulsion do not prominently appear.

Nonvolatile Solute

A nonvolatile solute is a solute that does not easily vaporize. This characteristic affects the vapor pressure of the solution significantly. When such a solute is dissolved into a solvent, it reduces the solvent's vapor pressure according to Raoult's Law. This is because the solute molecules occupy space at the liquid surface that would otherwise be occupied by solvent molecules, thus lowering the number of solvent molecules that can escape into the vapor phase. In calculations involving nonvolatile solutes, like in our example, the total vapor pressure of the solution is influenced solely by the solvent since the solute contributes no vapor pressure. This means all changes in pressure are attributed to the change in mole fractions of the solvent when the solute is added.