Problem 71 - General Chemistry: Principles and Modern Applications

Question

Adding \(1.00 \mathrm{g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) to \(80.00 \mathrm{g}\) cyclohexane, \(\mathrm{C}_{6} \mathrm{H}_{12},\) lowers the freezing point of the cyclohexane from 6.5 to \(3.3^{\circ} \mathrm{C}\). (a) What is the value of \(K_{f}\) for cyclohexane? (b) Which is the better solvent for molar mass determinations by freezing- point depression, benzene or cyclohexane? Explain.

Step-by-Step Solution

Answer

The value of \(K_{f}\) for cyclohexane is 20°C kg/mol. The solvent with the larger value of \(K_{f}\) would be a better solvent for molar mass determinations by freezing-point depression. Without the value of \(K_{f}\) for benzene, it can't be definitively said which solvent is better.

Step 1: Calculate the ΔTf

The depression in freezing point, ΔTf, is equal to the original freezing point of cyclohexane minus the new freezing point after the benzene has been added. Hence, ΔTf = 6.5°C - 3.3°C = 3.2°C.

Step 2: Convert grams of benzene to moles

On the other hand, the molecular weight of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), is 78.11 g/mol. Therefore, to get the number of moles, divide the given mass of benzene by the molecular weight. The result is 1.00g ÷ 78.11g/mol = 0.0128 mol benzene.

Step 3: Convert grams of cyclohexane to kilograms

We then convert the mass of the cyclohexane solvent to kilograms since molality is represented in terms of kilogram. Hence, 80.00g of cyclohexane = 80.00g ÷ 1000 = 0.0800 kg.

Step 4: Calculate the molality (m)

Having obtained the values of moles of solute and mass of solvent in kilograms, the molality can be calculated as: \( m = \frac{moles \, of \, solute}{mass \, of \, solvent \, (kg)} \) = \(\frac {0.0128}{0.0800}\) = 0.160 mol/kg.

Step 5: Calculate Kf

With the freezing point depression ΔTf = 3.2°C and the molality m = 0.16 mol/kg, the depression constant can be calculated using the formula ΔTf = \(K_f \times m\). Rearranging, \(K_f = \frac {ΔTf}{m}\) =\(\frac{3.2}{0.16}\) = 20°C kg/mol.

Step 6: Determine which is the better solvent

For molar mass determinations by freezing point depression, a better solvent would have a larger \(\(K_f\)\) value because a larger \(\(K_f\)\) would yield a larger change in freezing temperature for a given amount of solute. This makes it easier to measure the change in freezing point. Therefore, whichever between benzene and cyclohexane has the higher \(\(K_f\)\) will be considered the better solvent.

Key Concepts

Molality

Molality is an important concept in solution chemistry, particularly when dealing with colligative properties such as freezing point depression. It measures the concentration of a solute in a solution and is expressed as moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution and can change with temperature fluctuations, molality remains constant because it depends on the mass, not the volume, of the solvent. This makes it particularly useful in calculations involving temperature changes.
In the given exercise, we calculated the molality of the benzene-cyclohexane solution. First, we converted the mass of benzene into moles using its molecular weight. The result was 0.0128 moles. Then, we converted the mass of cyclohexane into kilograms, giving us 0.0800 kg. Using the formula for molality, \( m = \frac{moles \ of \ solute}{mass \ of \ solvent \ (kg)} \), we calculated the molality as 0.160 mol/kg. This value is crucial as it directly influences the extent of the freezing point depression in the solvent.

Freezing Point Constant

The freezing point constant, often denoted as \(K_f\), is a property of the solvent that measures its susceptibility to freezing point depression when a solute is added. Each solvent has a unique \(K_f\) value that depends on its chemical properties.
In the context of the exercise, we used the changed freezing point of cyclohexane upon adding benzene to determine its \(K_f\). The formula connecting freezing point depression, molality, and \(K_f\) is \( \Delta T_f = K_f \times m \), where \(\Delta T_f\) is the change in freezing point. Rearranging this formula, \(K_f = \frac{\Delta T_f}{m} \).

• Given \(\Delta T_f = 3.2^{\circ}C\) and \(m = 0.160\) mol/kg, substituting these values gives \(K_f = \frac{3.2}{0.160} = 20^{\circ}C \cdot kg/mol\).

This \(K_f\) value reveals that cyclohexane is quite sensitive to changes in its freezing point, making it a significant factor in deciding the appropriateness of a solvent for certain applications.

Solution Chemistry

Solution chemistry is the field of chemistry that studies the properties and behaviors of solutions, which are homogeneous mixtures of two or more substances. One of the intriguing aspects of solution chemistry is how solutes impact some physical properties of solvents, known as colligative properties. These include boiling point elevation, vapor pressure lowering, osmotic pressure, and freezing point depression.
Addressing further the aspect of freezing point depression, when a solute like benzene is dissolved in a solvent like cyclohexane, it disrupts the solid crystalline structure that the solvent molecules would form upon freezing. This leads to a lower freezing temperature for the solution than the pure solvent. The degree of these changes depends on the amount, not the type, of solute particles in the solution. As illustrated in the exercise, a small amount of benzene was enough to significantly change the freezing point of cyclohexane, clearly showing the practical implications of colligative properties.
This kind of solution manipulation is widely used in various fields, such as in road safety, where salt is spread on icy roads to lower the ice's melting point, thus keeping roads snow-free even in sub-zero temperatures. Understanding the principles of solution chemistry therefore provides a solid foundation for both academic insights and practical applications.