Problem 45 - 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 approximately 167.3 g/mol.

Step 1: Understand the Freezing Point Depression

The freezing point depression is a colligative property, which depends on the number of solute particles in the solution. The formula for freezing point depression is \[\Delta T_f = i \cdot K_f \cdot m\]Where \(\Delta T_f\) is the change in freezing point temperature, \(i\) is the van't Hoff factor (1 for a nonionic solute), \(K_f\) is the cryoscopic constant for water (1.86 \(\text{°C⋅kg/mol}\)), and \(m\) is the molality of the solution.

Step 2: Calculate the Change in Freezing Point

The solution's freezing point is given as \(-0.040^{\circ} \mathrm{C}\). The normal freezing point for water is \(0^{\circ} \mathrm{C}\). The change in freezing point \(\Delta T_f\) is then:\[\Delta T_f = 0 - (-0.040) = 0.040^{\circ} \mathrm{C}\].

Step 3: Calculate Molality of the Solution

Rearrange the freezing point depression formula to solve for molality, \(m\):\[m = \frac{\Delta T_f}{i \cdot K_f} = \frac{0.040}{1 \cdot 1.86} = 0.02151 \text{ mol/kg}\]The molality is approximately \(0.02151 \text{ mol/kg}\).

Step 4: Calculate the Moles of Solute

Using the calculated molality and the mass of the solvent, calculate the moles of solute:\[\text{moles} = \text{molality} \times \text{mass of water in kg} = 0.02151 \text{ mol/kg} \times 0.0500 \text{ kg} = 0.0010755 \text{ mol}\].

Step 5: Calculate Molar Mass of Solute

Finally, use the mass of the solute to calculate its molar mass:\[\text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{0.180 \text{ g}}{0.0010755 \text{ mol}} \approx 167.3 \text{ g/mol}\].

Key Concepts

Colligative Properties

Colligative properties are fascinating because they depend on the number of particles in a solution, not the particles' identity. This means that it doesn’t matter whether the solute is sugar or salt; what counts is how many particles are scattered in the solution. For colligative properties like freezing point depression or boiling point elevation:

• Adding more solute particles increases the effect. For instance, the more salt in water, the lower its freezing point.
• These properties are instrumental in determining characteristics like molar mass by observing changes in the solution's freezing or boiling points.

The beauty of colligative properties lies in their universality. This concept allows us to predict behavior in solutions without knowing the chemical nature of the solute.

Freezing Point Depression

Freezing point depression occurs when a solute is added to a solvent, causing the solvent's freezing point to drop. This process relies heavily on the simple yet crucial colligative principle that it’s all about the number, not the nature, of solute particles. Here's how it works:

• The regular arrangement of solvent molecules is disrupted by the solute particles, which makes it harder for the solvent to solidify at its normal freezing point.
• Thus, more energy (often removing heat) is required to draw the solvent into a solid state.

This comes in handy, especially in everyday applications like antifreeze in cars, which utilizes freezing point depression to prevent radiator water from freezing in cold weather.

Molality Calculation

Molality is a concentration unit that focuses on the moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of solution, molality remains unaffected by temperature changes. It's calculated by the formula:\[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\] Here's a simple breakdown:

• First, find the moles of solute either given directly or calculated using its mass and molar mass.
• Next, measure the mass of the solvent in kilograms.
• Use these to find the molality, which in turn can help calculate freezing point depression.

Molality is particularly useful in calculations involving colligative properties because it depends solely on mass, giving stable measurements regardless of thermodynamic changes.

Cryoscopic Constant

The cryoscopic constant (\(K_f\)) is a unique property of a solvent that quantifies the depression of its freezing point caused by solute particles. Every solvent has its own cryoscopic constant. For water, \(K_f\) is 1.86 °C·kg/mol. This constant plays a pivotal role in freezing point depression calculations, helping us determine:

• The extent to which the freezing point of the solvent is lowered.
• Calculations involving the determination of solute molar mass.

The cryoscopic constant is used in the formula: \[\Delta T_f = i \cdot K_f \cdot m\] This formula states that the change in freezing point (\(\Delta T_f\)) is a product of the cryoscopic constant, the van’t Hoff factor \(i\) (which counts how many particles the solute produces in solution), and the molality \(m\) of the solution. Understanding \(K_f\) thus helps in analyzing how the properties of a solution can transform with the addition of different solutes.