Question

The freezing point of a solution of \(2.40 \mathrm{~g}\) of biphenyl \(\left(\mathrm{C}_{12} \mathrm{H}_{10}\right)\) in \(75.0 \mathrm{~g}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(4.40^{\circ} \mathrm{C}\). The normal freezing point of benzene is \(5.50^{\circ} \mathrm{C}\). What is the molal freezing point constant \(\left({ }^{\circ} \mathrm{C} / \mathrm{m}\right)\) for benzene? (a) \(-5.3\) (b) \(-5.1\) (c) \(-4.6\) (d) \(-4.8\)

Short answer

The molal freezing point constant \(K_f\) for benzene is \(-5.1^\circ \mathrm{C} / \mathrm{m}\), which corresponds to option (b).

Step-by-step solution

Understand the concept of molal freezing point depression

Molal freezing point depression is a colligative property observed when the freezing point of a solution is lower than that of the pure solvent. The change in freezing point \(\Delta T_f\) is directly proportional to the molality \(m\) of the solution and is given by the formula \(\Delta T_f = K_f * m\), where \(K_f\) is the molal freezing point constant.

Calculate the change in freezing point \(\Delta T_f\)

The change in freezing point can be calculated by subtracting the freezing point of the solution from the normal freezing point of the pure solvent: \(\Delta T_f = T_{f,solvent} - T_{f,solution} = 5.50^\circ \mathrm{C} - 4.40^\circ \mathrm{C} = 1.10^\circ \mathrm{C}\).

Calculate the moles of solute (biphenyl)

Using the molar mass of biphenyl (\(C_{12}H_{10}\)) which is \(12 \times 12 + 10 \times 1 = 154 \mathrm{~g/mol}\), calculate the moles: \(moles_{solute} = \frac{mass_{solute}}{molar mass_{solute}} = \frac{2.40 \mathrm{~g}}{154 \mathrm{~g/mol}}\).

Calculate the molality (\(m\)) of the solution

Molality is moles of solute per kilogram of solvent. So, \(m = \frac{\moles_{solute}}{mass_{solvent} \mathrm{~in~kg}}\). After calculating the moles from Step 3, divide by the mass of solvent (75.0 g or 0.0750 kg) to find the molality.

Calculate the molal freezing point constant (\(K_f\))

Using the formula \(\Delta T_f = K_f * m\), rearrange to solve for \(K_f\): \[K_f = \frac{\Delta T_f}{m}\]. Use the values from Step 2 and Step 4.

Key concepts

Colligative Properties

Colligative properties are unique characteristics of solutions that depend on the number of solute particles dissolved in a solvent, regardless of their identity. This is an important concept in solution chemistry because it highlights how the addition of solute can alter the physical properties of a solution.

There are four main colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. When you dissolve a substance into a liquid, it disrupts the liquid's normal state, causing these changes in physical properties. For students, understanding these properties can be critical for explaining why, for example, salt is used to melt ice on roads (freezing point depression) or why adding antifreeze to a car radiator prevents the coolant inside from freezing (boiling point elevation).

An important note for students is that the effects observed through colligative properties are not dependent on what the solute is, but rather how much of it is present. That is, a 1 molal sugar solution would depress the freezing point of water by the same amount as a 1 molal sodium chloride solution, barring any ionic dissociation effects.

Freezing Point Depression

Freezing point depression is a colligative property that explains why adding a solute to a solvent decreases the temperature at which the solvent freezes. The more solute particles present, the lower the freezing point of the solution. This is because the solute particles interfere with the formation of the solvent's solid lattice structure.

The quantitative relationship between freezing point depression and the concentration of the solute in a solvent is given by the equation \( \Delta T_f = K_f \times m \) where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the molal freezing point constant of the solvent, and \( m \) is the molality of the solution. This equation is central to exercises where students need to calculate either the change in freezing point, the concentration of the solution, or the freezing point constant itself.

In the textbook exercise provided, the change in freezing point is calculated to be \( 1.10^\circ \text{C} \) by subtracting the freezing point of the solution from the normal freezing point of the solvent. This concept has practical applications, such as in the manufacturing of antifreeze and the formulation of freeze-resistant fluids.

Molality

Molality is a measure of concentration for a solution that is defined as the number of moles of solute per kilogram of solvent. This unit of concentration is preferred in certain contexts because it does not change with temperature, unlike molarity which varies due to the volume expansion or contraction of liquids with temperature.

Mathematically, molality (\(m\)) is calculated as follows: \(m = \frac{\text{moles of solute}}{\text{mass of solvent in kilograms}}\). To solve problems related to freezing point depression, it is crucial to calculate the molality accurately. This requires knowing the molar mass of the solute to determine the moles from the given mass, and the mass of the solvent must be accurately converted into kilograms.

For example, from the exercise provided, after you determine the moles of biphenyl, you divide by the mass of the benzene solvent in kilograms to get the molality of the solution. This step is fundamental before proceeding to calculate the molal freezing point constant, as it directly affects this value.

Solution Chemistry

Solution chemistry deals with the study of substances dissolved in a solvent to form a homogeneous mixture, known as a solution. It encompasses various aspects such as solubility, concentration, reaction rates in solution, and the properties of acids, bases, and salts when in solution.

The interplay of solute and solvent molecules is a cornerstone of solution chemistry and influences the colligative properties discussed earlier. A clear understanding of these interactions is necessary for anyone studying chemistry, as solutions are involved in many chemical processes and applications, from biological systems to industrial production.

In the context of the given exercise, we deal with a solution of biphenyl in benzene. Here, we focus on determining the molal freezing point constant (\( K_f \) of benzene), which is a unique physical constant for each solvent. It represents the degree to which the solvent's freezing point will be depressed per molal of solute and is essential for solving problems involving freezing point depression.