Problem 104 - Chemistry: The Central Science in SI Units, Global Edition

Question

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) boils at \(80.1^{\circ} \mathrm{C}\) and has a density of \(0.876 \mathrm{~g} / \mathrm{mL} .\) (a) When \(0.100 \mathrm{~mol}\) of a nondissociating solute is dissolved in \(500 \mathrm{~mL}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}\), the solution boils at \(79.52^{\circ} \mathrm{C}\). What is the molal boiling-point-elevation constant for \(\mathrm{C}_{6} \mathrm{H}_{6} ?\) (b) When \(10.0 \mathrm{~g}\) of a nondissociating unknown is dissolved in \(500 \mathrm{~mL}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}\), the solution boils at \(79.23^{\circ} \mathrm{C}\). What is the molar mass of the unknown?

Step-by-Step Solution

Answer

(a) The molal boiling-point-elevation constant (Kb) for benzene is -2.54 \(\frac{^{\circ}C \cdot kg}{mol}\). (b) The molar mass of the unknown solute is 66.5 g/mol.

Step 1: Part (a): Finding the molal boiling-point-elevation constant of benzene

First, we have to determine the boiling-point elevation (ΔTb) that happens when the solute is added. ΔTb is calculated by subtracting the normal boiling point of the solvent from the boiling point of the solution: ΔTb = T_solution - T_solvent = \(79.52^{\circ} C - 80.1^{\circ} C = -0.58^{\circ} C\) Next, we need to calculate the molality of the solute (m) which is given by the moles of solute divided by the mass of the solvent, expressed in kilograms: m = moles of solute / mass of solvent (in kg) Since the density of benzene is 0.876 g/mL, we can find the mass of 500 mL of benzene: Mass of benzene = \(0.876 \frac{g}{mL} \times 500 \, mL = 438 \, g\) Converting grams to kilograms, we get: Mass of benzene = \(438 \, g \times \frac{1 \, kg}{1000 \, g} = 0.438 \, kg\) Now, we are given 0.100 mol of solute, so the molality (m) is: m = \(\frac{0.100 \, mol}{0.438 \, kg} = 0.228 \frac{mol}{kg}\) Finally, we can find the molal boiling-point-elevation constant (Kb) for benzene using the boiling-point elevation equation: ΔTb = Kb × m Kb = \(\frac{ΔTb}{m} = \frac{-0.58^{\circ} C}{0.228 \frac{mol}{kg}} = -2.54 \frac{^{\circ}C \cdot kg}{mol}\)

Step 2: Part (b): Finding the molar mass of the unknown solute

Firstly, calculate the boiling-point elevation (ΔTb) for the solution with the unknown solute using the same procedure: ΔTb = T_solution - T_solvent = \(79.23^{\circ} C - 80.1^{\circ} C = -0.87^{\circ} C\) Now, we can rearrange the boiling-point elevation equation to find the molality (m) of the unknown solute: m = ΔTb / Kb = \(\frac{-0.87^{\circ} C}{-2.54 \frac{^{\circ}C \cdot kg}{mol}} = 0.343\frac{mol}{kg}\) We are given that 10.0 g of the unknown solute was dissolved in 500 mL of benzene. To find the molar mass (M) of the unknown solute, we need to find the mass of the solute in 1 kg of benzene. Since the mass of benzene is 0.438 kg, we can find the mass of the solute in 1 kg of benzene as: Mass of unknown in 1 kg of benzene = \(\frac{10.0 \, g}{0.438 \, kg} = 22.8 \, g\) Lastly, we use the molality to find the molar mass (M) of the unknown solute: M = Mass of unknown in 1 kg of benzene / m M = \(\frac{22.8 \, g}{0.343 \frac{mol}{kg}} = 66.5 \frac{g}{mol}\) So, the molar mass of the unknown solute is 66.5 g/mol.

Key Concepts

Molal Boiling-Point-Elevation Constant

When a solute is added to a solvent, the boiling point of the solution changes. This difference in boiling points is called boiling-point elevation. To measure how much a solvent's boiling point can be elevated by a solute, we use the molal boiling-point-elevation constant, denoted as \(K_b\). It is specific to each solvent.

The formula connecting these concepts is:

• \(\Delta T_b = K_b \times m\)

where \(\Delta T_b\) is the change in boiling point, \(K_b\) is the constant we need, and \(m\) is the molality of the solution.

In the exercise, the boiling-point decrease observed was \(-0.58^{\circ} C\) with a known solute molality. By plugging these into the formula, we found the \(K_b\) for benzene, which was \(-2.54 \frac{^{\circ} C \cdot kg}{mol}\). This constant helps us understand how benzene and other solvents behave when mixed with non-dissociating solutes.

Molality

Molality (\(m\)) is a measure of the concentration of a solute in a solution, expressed in moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality only considers the mass of the solvent, making it less sensitive to temperature fluctuations.

The formula for molality is:

• \(m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\)

In the exercise, to find the molality for benzene with a known solute, we used the mass of benzene in kilograms and the moles of solute. With 0.100 mol of solute and 0.438 kg of benzene, the molality calculated was \(0.228 \frac{mol}{kg}\).

Molality is particularly useful in boiling point elevation and freezing point depression calculations because it doesn’t change with temperature, providing consistent and reliable measurements.

Molecular Mass Determination

Establishing the molecular mass of an unknown compound when dissolved in a solvent involves both the boiling-point elevation principle and the molality concept. By measuring how the boiling point changes, we can work backwards to find the molality, then use it to determine molecular mass.

First, we measure the boiling-point change (\(\Delta T_b\)) and use the boiling-point-elevation constant \(K_b\) to find the molality of the unknown solute in the solution:

• \(m = \frac{\Delta T_b}{K_b}\)

Then, with known mass and volume of the solvent, we calculate the solute's mass per 1 kg of solvent. Finally, the molecular mass \(M\) of the solute is found using:

• \(M = \frac{\text{mass of solute in 1 kg of solvent}}{m}\)

In the exercise, this process resulted in a molecular mass of \(66.5 \frac{g}{mol}\) for the unknown compound, using benzene as the solvent and its specific \(K_b\). This method allows for precise molecular weight determination through simple boiling-point measurements.