Question

Calculate the molality and van't Hoff factor \((i)\) for the following aqueous solutions: (a) 1.00 mass \(\% \mathrm{NaCl}\), freezing point \(=-0.593^{\circ} \mathrm{C}\) (b) 0.500 mass \(\% \mathrm{CH}_{3} \mathrm{COOH}\), freezing point \(=-0.159^{\circ} \mathrm{C}\)

Short answer

NaCl: molality = 0.173 mol/kg, i = 1.83; CH₃COOH: molality = 0.0837 mol/kg, i = 1.02.

Step-by-step solution

- Define molality

Molality is defined as the number of moles of solute per kilogram of solvent. Use the formula: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]

- Molality of NaCl solution

For 1.00 mass \( \% \mathrm{NaCl} \), 1 g of \( \mathrm{NaCl}\) is present in 99 g of water. Calculate the moles of \(\mathrm{NaCl}\) as follows: \[ \text{moles of NaCl} = \frac{1.00 \ \text{g}}{58.44 \ \text{g/mol}} = 0.0171 \ \text{mol}\] Now convert the mass of water to kg: \[ 99 \ \text{g} \equiv 0.099 \ \text{kg} \] Finally, calculate the molality: \[ m = \frac{0.0171}{0.099} = 0.173 \ \text{mol/kg} \]

- Molality of CH₃COOH solution

For 0.500 mass \( \% \mathrm{CH}_{3} \mathrm{COOH} \), 0.500 g of \(\mathrm{CH}_{3} \mathrm{COOH} \) is present in 99.5 g of water. Calculate the moles of \( \mathrm{CH}_{3} \mathrm{COOH} \): \[ \text{moles of } \mathrm{CH}_{3} \mathrm{COOH} = \frac{0.500 \ \text{g}}{60.05 \ \text{g/mol}} = 0.00833 \ \text{mol} \] Convert the mass of water to kg: \[ 99.5 \ \text{g} \equiv 0.0995 \ \text{kg} \] Finally, calculate the molality: \[ m = \frac{0.00833}{0.0995} = 0.0837 \ \text{mol/kg} \]

- Calculate van't Hoff factor (i) for NaCl

Use the freezing point depression formula: \[ \Delta T_f = i \cdot K_f \cdot m \] For water, \Delta T_f = -0.593^{\circ} C and \ K_f = 1.86^{\circ} C \cdot \text{kg/mol}. Plug in the values: \[ 0.593 = i \cdot 1.86 \cdot 0.173 \] Solving for \ i: \[ i = \frac{0.593}{1.86 \cdot 0.173} = 1.83 \]

- Calculate van't Hoff factor (i) for CH₃COOH

Use the freezing point depression formula: \[ \Delta T_f = i \cdot K_f \cdot m \] For water, \Delta T_f = -0.159^{\circ} C and \ K_f = 1.86^{\circ} C \cdot \text{kg/mol}. Plug in the values: \[ 0.159 = i \cdot 1.86 \cdot 0.0837 \] Solving for \ i: \[ i = \frac{0.159}{1.86 \cdot 0.0837} = 1.02 \]

Key concepts

molality

Molality is a measure of the concentration of a solution. It is defined by the number of moles of solute divided by the mass (in kilograms) of the solvent. Unlike molarity, which measures the volume of the solution, molality uses the weight of the solvent. This makes it especially useful in scenarios involving temperature changes, since volume can fluctuate with temperature, but mass remains constant.
To calculate molality ( m ):
m = \( \frac{\text{moles of solute}}{\text{kg of solvent}} \).
For example, in a solution containing 1 g of NaCl in 99 g of water, first find the moles of NaCl: \( \frac{1.00 \text{g}}{58.44 \text{g/mol}} = 0.0171 \text{mol} \). Convert the mass of water to kg: 99 g = 0.099 kg.
Thus, the molality is: \( \frac{0.0171 \text{mol}}{0.099 \text{kg}} = 0.173 \text{mol/kg} \).

van't Hoff factor

The van't Hoff factor ( i ) is a measure of the effect of solute particles on colligative properties like boiling point elevation and freezing point depression. It represents the number of particles a solute separates into when dissolved.
For instance, NaCl dissociates into two ions: Na⁺ and Cl⁻. Therefore, its ideal van't Hoff factor is 2. However, real solutions often deviate from this ideal due to factors like ion pairing.
To calculate the van’t Hoff factor using freezing point depression, use the equation: \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant of the solvent, and \( m \) is the molality of the solution.
For a NaCl solution with \( \Delta T_f = -0.593^{\circ}C \), \( K_f = 1.86^{\circ}C \cdot \text{kg/mol} \), and \( m = 0.173 \text{mol/kg} \):
Solve for i : \( i = \frac{0.593}{1.86 \cdot 0.173} = 1.83 \).

freezing point depression

Freezing point depression occurs when a solute is added to a solvent, lowering the temperature at which the solvent freezes. This is a colligative property, meaning it depends on the number of solute particles, not their identity.
The formula for freezing point depression is: \( \Delta T_f = i \cdot K_f \cdot m \), where:\- \( \Delta T_f \) is the decrease in freezing point. \- \( i \) is the van’t Hoff factor, indicating the number of particles into which the solute dissociates. \- \( K_f \) is the cryoscopic constant of the solvent. \- \( m \) is the molality of the solution.
Using this formula, one can calculate the difference in freezing points between pure solvents and their solutions, aiding in determining the van’t Hoff factor and hence the behavior of the solute in the solution.
For CH₃COOH with a known \( \Delta T_f \) of -0.159°C, \( K_f = 1.86^{\circ}C \cdot \text{kg/mol} \), and \( m = 0.0837 \text{mol/kg} \), we find \( i = \frac{0.159}{1.86 \cdot 0.0837} = 1.02 \), indicating almost no dissociation.

aqueous solutions

Aqueous solutions are those in which the solvent is water. They are extremely common in chemistry due to water's ability to dissolve many substances. The properties of aqueous solutions depend on the solute and the interactions between the solute and water. Key colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure are affected.
In our example, NaCl in water illustrates a simple ion-dissociation, leading to a lowering of the freezing point.
Similarly, a weak acid like acetic acid (CH₃COOH) in water partially dissociates, affecting the freezing point to a lesser degree than a strong electrolyte like NaCl.
These behaviors are quantified using molality and the van't Hoff factor, aiding in understanding solute dynamics in aqueous solutions.

solute and solvent interactions

The interactions between solute and solvent greatly influence the properties of a solution. For instance, ionic solutes like NaCl dissociate into ions in water, resulting in significant changes in properties like freezing point. This interaction is measured by changes in colligative properties and the van't Hoff factor.
For NaCl, strong ion-dipole interactions with water lead to substantial freezing point depression, allowing calculation of its van’t Hoff factor and behavior in solution.
On the other hand, molecular solutes like CH₃COOH may not dissociate completely. As a weak acid, its partial ionization in water results in a smaller impact on freezing point depression, reflected in a van't Hoff factor close to 1.
Understanding these interactions helps predict how different solutes will affect solution properties, critical in applications ranging from chemistry to various practical processes.