Question

Calculate the freezing point and the boiling point of each of the following solutions. (Assume complete dissociation.) a. 5.0 \(\mathrm{g} \mathrm{NaCl}\) in 25 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) b. 2.0 \(\mathrm{g} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) in 15 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\)

Short answer

The freezing and boiling points of the solutions are as follows: a. For the 5.0 g NaCl solution in 25 g H2O: Freezing point: \(-12.7\text{ °C}\) and boiling point: \(103.5\text{ °C}\) b. For the 2.0 g Al(NO₃)₃ solution in 15 g H2O: Freezing point: \(-4.66\text{ °C}\) and boiling point: \(101.3\text{ °C}\)

Step-by-step solution

Calculate the molality of the NaCl solution

To calculate the molality of the NaCl solution, we will first convert the given solute and solvent masses to moles and kilograms: Moles of NaCl = \(\frac{5\text{g}}{58.44\text{g/mol}}\) = 0.0856 mol (using the molar mass of NaCl) Kilograms of water = \(\frac{25\text{g}}{1000\text{g/kg}}\) = 0.025 kg Now, we can calculate the molality: Molality (m) = \(\frac{0.0856\text{ mol}}{0.025\text{ kg}}\) = 3.42 mol/kg

Calculate the freezing point depression and boiling point elevation for NaCl solution

To calculate the freezing point depression and boiling point elevation for the NaCl solution, we will use the formulas above. Since NaCl is a strong electrolyte and dissociates into two ions (Na+ and Cl-), the van't Hoff factor (i) is 2: Freezing point depression: \(ΔT_f = K_f × m × i = 1.86\text{ °C/mol·kg} × 3.42\text{ mol/kg} × 2 = 12.7\text{ °C}\) Boiling point elevation: \(ΔT_b = K_b × m × i = 0.512\text{ °C/mol·kg} × 3.42\text{ mol/kg} × 2 = 3.50\text{ °C}\)

Calculate the new freezing and boiling points of the NaCl solution

Now, we can calculate the new freezing and boiling points by subtracting the freezing point depression and adding the boiling point elevation to the normal freezing and boiling points of water (0°C and 100°C, respectively): New freezing point = \(0\text{ °C} - 12.7\text{ °C} = -12.7\text{ °C}\) New boiling point = \(100\text{ °C} + 3.50\text{ °C} = 103.5\text{ °C}\) So, the freezing and boiling points of the 5.0 g NaCl solution in 25 g H2O are -12.7°C and 103.5°C, respectively. ##Solution B##

Calculate the molality of the Al(NO3)3 solution

To calculate the molality of the Al(NO₃)₃ solution, we will first convert the given solute and solvent masses to moles and kilograms: Moles of Al(NO₃)₃ = \(\frac{2\text{g}}{213.0\text{g/mol}}\) = 0.00939 mol (using the molar mass of Al(NO₃)₃) Kilograms of water = \(\frac{15\text{g}}{1000\text{g/kg}}\) = 0.015 kg Now, we can calculate the molality: Molality (m) = \(\frac{0.00939\text{ mol}}{0.015\text{ kg}}\) = 0.626 mol/kg

Calculate the freezing point depression and boiling point elevation for Al(NO3)3 solution

To calculate the freezing point depression and boiling point elevation for the Al(NO₃)₃ solution, we will use the same formulas. Since Al(NO₃)₃ is a strong electrolyte and dissociates into four ions (Al³+ and 3 NO₃⁻), the van't Hoff factor (i) is 4: Freezing point depression: \(ΔT_f = K_f × m × i = 1.86\text{ °C/mol·kg} × 0.626\text{ mol/kg} × 4 = 4.66\text{ °C}\) Boiling point elevation: \(ΔT_b = K_b × m × i = 0.512\text{ °C/mol·kg} × 0.626\text{ mol/kg} × 4 = 1.28\text{ °C}\)

Calculate the new freezing and boiling points of the Al(NO3)3 solution

Now, we can calculate the new freezing and boiling points by subtracting the freezing point depression and adding the boiling point elevation to the normal freezing and boiling points of water (0°C and 100°C, respectively): New freezing point = \(0\text{ °C} - 4.66\text{ °C} = -4.66\text{ °C}\) New boiling point = \(100\text{ °C} + 1.28\text{ °C} = 101.3\text{ °C}\) So, the freezing and boiling points of the 2.0 g Al(NO₃)₃ solution in 15 g H2O are -4.66°C and 101.3°C, respectively.

Key concepts

Molality

Molality 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, which depends on the volume of the solution, molality is solely based on mass. This makes it especially useful in situations where temperature changes, as it remains constant.
Calculating molality involves two primary steps: determining the number of moles of solute and the mass of the solvent in kilograms. For example, with 5.0 g of NaCl dissolved in 25 g of water, we first convert 5.0 g of NaCl to moles using its molar mass (58.44 g/mol). This gives us approximately 0.0856 moles. Next, we convert the 25 g of water to kilograms, resulting in 0.025 kg.
We then calculate the molality by dividing the moles of solute by the kilograms of solvent, which yields the concentration of the solution in mol/kg. Understanding molality is crucial for dealing with colligative properties, as it plays a major role in equations used for calculating changes in the boiling and freezing points of solutions.

Freezing Point Depression

Freezing point depression refers to the decrease in the freezing point of a solvent when a solute is added. This phenomenon is a colligative property, meaning it depends on the number of particles in a solution, not on their identity.
For a solution like NaCl in water, the introduction of solute particles disrupts the ability of solvent molecules (water) to form a solid structure at the normal freezing point. This disruption causes the solution to freeze at a lower temperature.
The extent of freezing point depression is calculated using the formula: \(\Delta T_f = K_f \times m \times i\). Here, \(\Delta T_f\) is the change in freezing point, \(K_f\) is the cryoscopic constant specific to the solvent, \(m\) is the molality, and \(i\), the van't Hoff factor, is the number of particles into which the solute dissociates.
For water, \(K_f\) is typically given as 1.86 \(^{\circ}\)C/mol·kg. In the case of NaCl, which dissociates completely into two ions, the van't Hoff factor is 2. Thus, the formula considers both the concentration and the dissociation of solute particles.

Boiling Point Elevation

Boiling point elevation is the increase in the boiling point of a solvent due to the addition of a solute. Like freezing point depression, it is another colligative property.
The presence of solute particles in a solution requires the solvent to reach a higher temperature to achieve the vapor pressure necessary to boil. This results in a higher boiling point for the solution compared to the pure solvent.
To calculate the boiling point elevation, we use the equation: \(\Delta T_b = K_b \times m \times i\). Here, \(\Delta T_b\) represents the increase in boiling temperature, \(K_b\) is the ebullioscopic constant of the solvent, \(m\) is the molality, and \(i\) is the van't Hoff factor.
For water, \(K_b\) is often approximated as 0.512 \(^{\circ}\)C/mol·kg. This calculation considers both the concentration of the solution and the outcome of the solute's dissociation, further affecting the extent to which the boiling point is elevated.

Electrolyte Dissociation

Electrolyte dissociation is the process in which ionic compounds break into their constituent ions in a solution. When an electrolyte, such as NaCl, dissolves in water, it dissociates into Na⁺ and Cl⁻ ions, creating a conductive solution.
This behavior is essential in colligative properties because the number of particles produced directly influences calculations for freezing point depression and boiling point elevation.
The more ions produced from dissociation, the greater the effect on the solution's colligative properties. For example, NaCl produces two ions, while a compound like Al(NO₃)₃ dissociates into four ions (one Al³⁺ ion and three NO₃⁻ ions).
Understanding the extent of dissociation helps to accurately calculate the total effect of a solute on the properties of a solution, especially for strong electrolytes that dissociate completely. In our calculations, this dissociation is accounted for with the van't Hoff factor.

Van't Hoff Factor

The van't Hoff factor, denoted as \(i\), is a measure of the degree of dissociation of a solute in solution. It represents the number of particles a solute provides upon dissociation.
In colligative property calculations, \(i\) is crucial because it directly affects how solute concentration is utilized in formulas for both freezing point depression and boiling point elevation.
For instance, in the case of NaCl, complete dissociation results in two particles (Na⁺ and Cl⁻), so \(i = 2\). For Al(NO₃)₃, the \(i = 4\), because it dissociates into one Al³⁺ and three NO₃⁻ ions.
The van't Hoff factor helps adjust molality to account for the actual number of solute particles in the solution, thus making it possible to accurately calculate the changes in colligative properties. Careful consideration of \(i\) ensures precise experimental and theoretical predictions of solution behavior.