Question

Calculate the freezing point and normal boiling points of each of the following aqueous solutions. (a) \(2.63 \mathrm{~m}\) acetic acid (b) \(33.0 \%\) by mass lactose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (c) \(32.15 \mathrm{~mL}\) of ethylene glycol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}(d=1.113 \mathrm{~g} / \mathrm{mL})\) in \(624 \mathrm{~mL}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\)

Short answer

Question: Calculate the freezing point and normal boiling points of three aqueous solutions: acetic acid, lactose, and ethylene glycol. Answer: The freezing point and normal boiling points of the three aqueous solutions are as follows: 1. Acetic acid solution: - Freezing point: -4.88 °C - Boiling point: 101.35 °C 2. Lactose solution: - Freezing point: -2.66 °C - Boiling point: 100.73 °C 3. Ethylene glycol solution: - Freezing point: -1.72 °C - Boiling point: 100.47 °C

Step-by-step solution

(Find the molality of acetic acid)

The molality, \(m\), of acetic acid is given as \(2.63 \ \text{mol/kg}\). Since the molality is already known, we can directly use the formulas for freezing point depression,ΔT_f= K_f \cdot m\(, and boiling point elevation, \)\Delta T_b = K_b \cdot m$.

(Calculate freezing point and boiling point of acetic acid solution)

Firstly, we will calculate the freezing point. The freezing point depression constant for water is \(K_f = 1.86\, ^\circ\text{C/m}\). We will use the provided molality of acetic acid: \(\Delta T_f = K_f \cdot m = 1.86\, ^\circ\text{C/m} \cdot 2.63\, \text{mol/kg} = 4.88^\circ\text{C}\) The freezing point of the acetic acid solution is \(0 ^\circ\text{C} - 4.88 ^\circ\text{C} = -4.88 ^\circ\text{C}\). Now, let's calculate the boiling point. The boiling point elevation constant for water is \(K_b = 0.512\, ^\circ\text{C/m}\). Using the provided molality of acetic acid: \(\Delta T_b = K_b \cdot m = 0.512\, ^\circ\text{C/m} \cdot 2.63\, \text{mol/kg} = 1.35 ^\circ\text{C}\) The boiling point of the acetic acid solution is \(100 ^\circ\text{C} + 1.35 ^\circ\text{C} = 101.35 ^\circ\text{C}\).

(Find the molality of lactose solution)

In this case, we are given that the mass percentage of lactose in the solution is \(33.0\%\). To find the molality, we need to find the mass of lactose and mass of water in a \(100\, \text{g}\) sample of the solution. We have: Mass of lactose = \(0.330 \times 100 \ \text{g} = 33 \ \text{g}\) Mass of water = \(100 - 33 \ \text{g} = 67 \ \text{g}\) We can convert the masses into moles: Moles of lactose = \(\frac{33\, \text{g}}{342.3\, \text{g/mol}} = 0.096\, \text{mol}\) (molar mass of \(\text{C}_{12}\text{H}_{22}\text{O}_{11} = 342.3\, \text{g/mol}\)) Now we will find the molality, \(m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}\): Molality of lactose solution = \(\frac{0.096 \ \text{mol}}{0.067\, \text{kg}} = 1.43 \ \text{mol/kg}\)

(Calculate freezing point and boiling point of lactose solution)

We will use the molality of lactose in the formulas for freezing point depression and boiling point elevation, as we did before: \(\Delta T_f = K_f \cdot m = 1.86\, ^\circ\text{C/m} \cdot 1.43\, \text{mol/kg} = 2.66 ^\circ\text{C}\) Freezing point of lactose solution = \(0 ^\circ\text{C} - 2.66 ^\circ\text{C} = -2.66 ^\circ\text{C}\) \(\Delta T_b = K_b \cdot m = 0.512\, ^\circ\text{C/m} \cdot 1.43\, \text{mol/kg} = 0.73 ^\circ\text{C}\) Boiling point of lactose solution = \(100 ^\circ\text{C} + 0.73 ^\circ\text{C} = 100.73 ^\circ\text{C}\)

(Find the molality of ethylene glycol solution)

We are given the volume of ethylene glycol as \(32.15\, \text{mL}\) and its density as \(1.113\, \text{g/mL}\). To find the molality, we will first find the mass of ethylene glycol and mass of water: Mass of ethylene glycol = \(32.15\, \text{mL} \times 1.113 \ \text{g/mL} = 35.8\, \text{g}\) Mass of water = \(624\, \text{mL} \times 1.00\, \text{g/mL} = 624\, \text{g}\) We can convert the masses into moles: Moles of ethylene glycol = \(\frac{35.8\, \text{g}}{62.07\, \text{g/mol}} = 0.577\, \text{mol}\) (molar mass of \(\text{C}_2\text{H}_6\text{O}_2 = 62.07\, \text{g/mol}\)) Now we will find the molality of ethylene glycol solution: Molality of ethylene glycol solution = \(\frac{0.577\, \text{mol}}{0.624\, \text{kg}} = 0.925 \ \text{mol/kg}\)

(Calculate freezing point and boiling point of ethylene glycol solution)

Now, we will use the molality of ethylene glycol in the formulas for freezing point depression and boiling point elevation: \(\Delta T_f = K_f \cdot m = 1.86\, ^\circ\text{C/m} \cdot 0.925\, \text{mol/kg} = 1.72 ^\circ\text{C}\) Freezing point of ethylene glycol solution = \(0 ^\circ\text{C} - 1.72 ^\circ\text{C} = -1.72 ^\circ\text{C}\) \(\Delta T_b = K_b \cdot m = 0.512\, ^\circ\text{C/m} \cdot 0.925\, \text{mol/kg} = 0.47 ^\circ\text{C}\) Boiling point of ethylene glycol solution = \(100 ^\circ\text{C} + 0.47 ^\circ\text{C} = 100.47 ^\circ\text{C}\)

Key concepts

Freezing Point Depression

The phenomenon where the presence of a solute causes a reduction in the freezing point of a solution compared to that of the pure solvent is known as freezing point depression. It is one of the colligative properties, which means it's dependent on the quantity, not the identity, of the solute particles.

In practice, this effect can be calculated using the equation:
\[\begin{equation} \(\Delta T_f = K_f \cdot m\)\end{equation}\]where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the freezing point depression constant, specific to each solvent, and \(m\) is the molality of the solution.

This concept is applied in many real-world scenarios, such as by adding salt to ice on the roads during winter, which lowers the freezing point and prevents ice formation.

Boiling Point Elevation

Boiling point elevation refers to the physical fact that a solution has a higher boiling point than the pure solvent. Just like freezing point depression, boiling point elevation is a colligative property, thus it only depends on the amount of dissolved particles and not on their chemical nature.

The formula used to calculate the boiling point elevation is very similar to that for freezing point depression:
\[\begin{equation} \(\Delta T_b = K_b \cdot m\)\end{equation}\]where \(\Delta T_b\) is the boiling point elevation, \(K_b\) is the boiling point elevation constant, and \(m\) is the molality of the solution. This principle is applied in cooking, where salt is added to water to increase the boiling point and potentially cook food at a higher temperature.

Molality

Molality is a measure of the concentration of a solution, expressed as the number of moles of solute per kilogram of solvent. It differs from molarity, which is moles of solute per liter of solution. Its formula is: \[\begin{equation} \(m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}\)\end{equation}\]Molality is particularly useful in calculations involving changes in physical properties of solutions, such as boiling point elevation and freezing point depression, because unlike molarity, molality is independent of changes in temperature and pressure.

Solution Concentration

Solution concentration is a quantitative measure of the amount of solute in a given amount of solvent or solution. There are various ways to express solution concentration, such as molality, molarity, mass percent, and volume percent. The choice of concentration measure depends on the scenario and the precise nature of the solution in question.

In the context of colligative properties, molality is often most appropriate, as it isn't affected by temperature changes and provides a direct link to the number of solute particles in the solvent, which in turn affects properties like freezing point and boiling point.