Question

The freezing point of benzene is \(5.5^{\circ} \mathrm{C}\). What is the freezing point of a solution of \(5.00 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) in \(444 \mathrm{~g}\) of benzene \(\left(K_{\mathrm{f}}\right.\) of benzene \(\left.=4.90^{\circ} \mathrm{C} / \mathrm{m}\right) ?\)

Short answer

5.07^{\circ}C

Step-by-step solution

Determine the molality (m)

First, find the number of moles of naphthalene (C_{10}H_{8}). The molar mass of naphthalene is calculated as: \[ 12 \times 10 + 1 \times 8 = 128 \text{ g/mol} \] Next, calculate the number of moles (n) of naphthalene: \[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \text{ g}}{128 \text{ g/mol}} = 0.0391 \text{ mol} \] Now, find the molality (m), which is moles of solute per kilogram of solvent: \[ m = \frac{0.0391 \text{ mol}}{0.444 \text{ kg}} = 0.0881 \text{ m} \]

Calculate the freezing point depression \(\Delta T_f\)

Use the formula for freezing point depression: \[ \Delta T_f = K_f \times m \] where \( K_f = 4.90^{\circ}C/m \) and \( m = 0.0881 \). So, \( \Delta T_f \) is: \[ \Delta T_f = 4.90^{\circ}C/m \times 0.0881 \, m = 0.4317^{\circ}C \]

Determine the new freezing point of the solution

Subtract the freezing point depression from the pure solvent's freezing point: \[ T_f(\text{solution}) = T_f(\text{pure benzene}) - \Delta T_f \] Benzene's freezing point is \(5.5^{\circ} \text{C} \), so: \[ T_f(\text{solution}) = 5.5^{\circ} C - 0.4317^{\circ}C = 5.0683^{\circ}C \]

Key concepts

Molality

Molality is a measure of the concentration of a solute in a solution. It's different from molarity, as it focuses on the moles of solute per kilogram of solvent rather than per liter of solution.
To find molality in the given problem, we used the formula:
\( m = \frac{\text{moles of solute}}{\text{kilogram of solvent}} \)
First, we calculated the moles of naphthalene using its molar mass. Next, we divided this value by the mass of the benzene solvent in kilograms. Understanding molality is key in figuring out how strong a solution is, which directly impacts properties like freezing point depression.

Molar Mass

Molar mass is the mass of one mole of a substance. It's calculated by summing the atomic masses of all atoms in a molecule. For naphthalene (\( C_{10}H_{8} \)), its molar mass is:
\(12 \text{ (carbon)} \times 10 + 1 \text{ (hydrogen)} \times 8 = 128 \text{ g/mol} \)
This unit (grams per mole) helps in converting the mass of a substance to moles, which is crucial in various chemical calculations, including those involving colligative properties like freezing point depression.

Colligative Properties

Colligative properties depend only on the number of solute particles in a solution, not their identity. Examples include boiling point elevation, vapor pressure lowering, osmotic pressure, and freezing point depression.
In this exercise, we focused on freezing point depression, where the addition of a solute lowers the freezing point of the solvent. The magnitude is determined using:\( \Delta T_f = K_f \times m \)
Where \( K_f \)is the cryoscopic constant, and \( m \)is the molality.
Understanding these properties helps predict how a solute will affect the physical properties of a solution, which is vital in various real-world applications such as antifreeze in car engines.

Solution Chemistry

Solution chemistry studies how substances dissolve and interact in solvents. When naphthalene dissolves in benzene, it forms a solution where the properties of the mixture differ from the pure solvents.
Knowledge of solution chemistry allows us to predict how solutes will behave, how concentrated a solution can get, and how its properties will change. This is foundational for solving problems involving colligative properties and is crucial in both laboratory settings and industrial applications.
Understanding these interactions at a molecular level helps in designing experiments and industrial processes efficiently.