Problem 43 - Chemistry and Chemical Reactivity

Question

An aqueous solution contains \(0.180 \mathrm{g}\) of an unknown, nonionic solute in \(50.0 \mathrm{g}\) of water. The solution freezes at \(-0.040^{\circ} \mathrm{C} .\) What is the molar mass of the solute?

Step-by-Step Solution

Answer

The molar mass of the solute is 167.44 g/mol.

Step 1: Understand the freezing point depression

Freezing point depression is represented by the formula \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( K_f \) is the cryoscopic constant of the solvent (for water, \( K_f = 1.86 \, ^\circ\mathrm{C}\cdot \mathrm{kg/mol} \)), and \( m \) is the molality of the solution.

Step 2: Calculate the change in freezing temperature

The freezing point of pure water is \( 0.0 \, ^\circ\mathrm{C} \). The solution freezes at \( -0.040 \, ^\circ\mathrm{C} \). Thus, the change in freezing temperature, \( \Delta T_f \), is \(-0.040 \, ^\circ\mathrm{C} - 0 \, ^\circ\mathrm{C} = -0.040 \, ^\circ\mathrm{C} \).

Step 3: Calculate molality of the solution

Using the formula \( \Delta T_f = K_f \cdot m \), rearrange to solve for \( m \):\[m = \frac{\Delta T_f}{K_f} = \frac{0.040}{1.86} = 0.0215 \, \mathrm{mol/kg}\].

Step 4: Calculate moles of solute

The molality \( m \) is equal to the moles of solute per kilogram of solvent. If \( m = 0.0215 \, \mathrm{mol/kg} \) and the solvent's mass is \( 0.050 \, \mathrm{kg} \) (since 50 g of water is used), the moles of solute is \( 0.0215 \, \mathrm{mol/kg} \times 0.050 \, \mathrm{kg} = 0.001075 \, \mathrm{mol} \).

Step 5: Determine the molar mass of the solute

Molar mass is calculated using the equation: \(\text{Molar mass} = \frac{\text{mass of solute in grams}}{\text{moles of solute}} \).Substitute the given data: \( \frac{0.180 \, \mathrm{g}}{0.001075 \, \mathrm{mol}} = 167.44 \, \mathrm{g/mol} \).

Key Concepts

Molar mass calculation

Molar mass is a fundamental concept in chemistry that helps us understand the mass of a given substance's molecules. When calculating molar mass, it's important to realize that it refers to the mass of one mole of a substance. In practical terms, molar mass is calculated as the mass of the solute, typically measured in grams, divided by the number of moles of solute. For example, if you know the number of moles and the mass of the solute, you can readily find the molar mass. Given a mass of solute (in this case, 0.180 g) and the calculated moles of solute (0.001075 mol), the molar mass is determined using the formula:

• Molar mass = \( \frac{\text{mass of solute in grams}}{\text{moles of solute}} \)
• For this specific problem, the molar mass is \( \frac{0.180 \text{ g}}{0.001075 \text{ mol}} = 167.44 \text{ g/mol} \)

Understanding molar mass can provide insights into the nature and properties of the substance being analyzed.

Cryoscopic constant

The cryoscopic constant, symbolized as \( K_f \), is a property-specific constant that varies according to the solvent being used. It directly relates to how much the freezing point of a solvent will decrease when a non-volatile solute is added. This is a key variable in the formula for freezing point depression \( \Delta T_f = K_f \cdot m \).

• The cryoscopic constant of water is \( 1.86 \, ^\circ\mathrm{C}\cdot \mathrm{kg/mol} \).
• This means that the addition of 1 mol of solute to 1 kg of water will lower the freezing point by 1.86 degrees Celsius.

Knowing \( K_f \) is important for calculations involving colligative properties such as freezing point depression.

Molality

Molality, denoted as \( m \), is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality does not change with temperature since it does not involve the volume of the solution. Here's how it is calculated:

• Rearrange the freezing point depression equation \( \Delta T_f = K_f \cdot m \) to solve for \( m \).
• Use the change in freezing point \( \Delta T_f \) of -0.040°C and the known \( K_f \) for water.
• Molality is \( \frac{0.040}{1.86} = 0.0215 \, \mathrm{mol/kg} \).

This calculated molality helps in determining the number of moles of solute when given the mass of the solvent.

Nonionic solute

A nonionic solute is a substance that does not dissociate into ions when dissolved in a solvent. This type of solute doesn't carry a charge and hence doesn't alter the electrical conductivity of the solution. Understanding nonionic solutes can clarify why certain solutions behave as they do under various conditions.

• Nonionic solutes affect the physical properties of solvents, like freezing point depression, in predictable ways without contributing to electrical conduction.
• For calculations like freezing point depression, the presence of a nonionic solute results in a straightforward application of formulas since there are no ion-dissociation factors to consider.
• This simplicity allows for easier determination of properties such as molar mass.

Recognizing the nature of a solute being nonionic streamlines understanding the related mathematical problems.