Problem 67 - Chemistry The Molecular Science

Question

Calculate the boiling point and the freezing point of these solutions at \(760 \mathrm{mmHg}\). (a) \(20.0 \mathrm{~g}\) citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), in \(100.0 \mathrm{~g}\) water (b) \(3.00 \mathrm{~g} \mathrm{CH}_{3} \mathrm{I}\) in \(20.0 \mathrm{~g}\) benzene \(\left(K_{\mathrm{b}}\right.\) benzene \(=\) \(2.53^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol} ; K_{\mathrm{f}} \text { benzene } \left.=-5.10^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\right)\)

Step-by-Step Solution

Answer

The freezing point of the citric acid solution is \(-1.93^{\circ}C\) and boiling point is \(100.532^{\circ}C\). For methyl iodide in benzene, freezing point is \(0.12^{\circ}C\) and boiling point is \(82.77^{\circ}C\).

Step 1: Calculate molality for citric acid

To determine the molality of the solution, we first need to find the number of moles of citric acid and mass of the solvent in kg. The molar mass of citric acid \( (\mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_7) \) is \(192.13 \mathrm{~g/mol}\). Number of moles \( = \frac{20.0 \mathrm{~g}}{192.13 \mathrm{~g/mol}} = 0.104 \mathrm{~mol}\). Mass of water \( = 100.0 \mathrm{~g} = 0.100 \mathrm{~kg}\). Molality (\( m \)) is calculated as:\[ m = \frac{0.104 \mathrm{~mol}}{0.100 \mathrm{~kg}} = 1.04 \mathrm{~mol/kg} \]

Step 2: Calculate freezing point depression for citric acid solution

The freezing point depression formula is \[ \Delta T_f = i \cdot K_f \cdot m \] where \( i \) is the van 't Hoff factor (typically 1 for non-electrolytes like citric acid in dilute aqueous solutions), \( K_f \) is the freezing point depression constant for water (\(-1.86^{\circ} \mathrm{C} \mathrm{kg/mol}\)), and \( m \) is the molality.\[ \Delta T_f = 1 \times (-1.86^{\circ} \mathrm{C} \mathrm{kg/mol}) \times 1.04 \mathrm{~mol/kg} \approx -1.93^{\circ} \mathrm{C} \]The normal freezing point of water is \(0^{\circ} \mathrm{C}\), so the freezing point of the solution is \[0^{\circ} \mathrm{C} - 1.93^{\circ} \mathrm{C} = -1.93^{\circ} \mathrm{C}\]

Step 3: Calculate boiling point elevation for citric acid solution

The boiling point elevation formula is \[ \Delta T_b = i \cdot K_b \cdot m \]where \( i = 1 \), \( K_b \) for water is \(0.512^{\circ} \mathrm{C} \mathrm{kg/mol}\), and \( m \) is the molality.\[ \Delta T_b = 1 \times 0.512^{\circ} \mathrm{C} \mathrm{kg/mol} \times 1.04 \mathrm{~mol/kg} \approx 0.532^{\circ} \mathrm{C} \]The normal boiling point of water is \(100^{\circ} \mathrm{C}\), so the boiling point of the solution is\[100^{\circ} \mathrm{C} + 0.532^{\circ} \mathrm{C} = 100.532^{\circ} \mathrm{C}\]

Step 4: Calculate molality for methyl iodide

Find the molality of methyl iodide (\(\mathrm{CH}_3\mathrm{I}\)) in benzene. The molar mass of \(\mathrm{CH}_3\mathrm{I}\) is \(141.94 \mathrm{~g/mol}\).Number of moles \(= \frac{3.00 \mathrm{~g}}{141.94 \mathrm{~g/mol}} = 0.0211 \mathrm{~mol}\).Mass of benzene \(= 20.0 \mathrm{~g} = 0.0200 \mathrm{~kg}\).Molality \( m = \frac{0.0211 \mathrm{~mol}}{0.0200 \mathrm{~kg}} = 1.055 \mathrm{~mol/kg} \) approximately.

Step 5: Calculate freezing point depression for methyl iodide solution

Using the freezing point depression formula \[ \Delta T_f = i \cdot K_f \cdot m \]where \( i = 1 \), \( K_f \) for benzene \(=-5.10^{\circ} \mathrm{C} \mathrm{kg/mol}\), and \( m = 1.055 \mathrm{~mol/kg}\).\[ \Delta T_f = 1 \times (-5.10^{\circ} \mathrm{C} \mathrm{kg/mol}) \times 1.055 \mathrm{~mol/kg} \approx -5.38^{\circ} \mathrm{C} \] Since benzene freezes at \(5.5^{\circ} \mathrm{C}\), the solution's freezing point is: \[ 5.5^{\circ} \mathrm{C} - 5.38^{\circ} \mathrm{C} = 0.12^{\circ} \mathrm{C} \]

Step 6: Calculate boiling point elevation for methyl iodide solution

The boiling point elevation is calculated by \[ \Delta T_b = i \cdot K_b \cdot m \]where \( i = 1 \), \( K_b \) for benzene \(= 2.53^{\circ} \mathrm{C} \mathrm{kg/mol}\), and \( m = 1.055 \mathrm{~mol/kg}\).\[ \Delta T_b = 1 \times 2.53^{\circ} \mathrm{C} \mathrm{kg/mol} \times 1.055 \mathrm{~mol/kg} \approx 2.67^{\circ} \mathrm{C} \]Since benzene boils at \(80.1^{\circ} \mathrm{C}\), the boiling point of the solution is:\[ 80.1^{\circ} \mathrm{C} + 2.67^{\circ} \mathrm{C} = 82.77^{\circ} \mathrm{C} \]

Key Concepts

Boiling Point Elevation

Boiling point elevation is a colligative property that depends on the number of solute particles in a solution, not the type of particles. When a non-volatile solute is added to a liquid, it elevates the boiling point of the solvent. The change in boiling point (\(\Delta T_b\)) is calculated using the formula:

• \(\Delta T_b = i \cdot K_b \cdot m\)

Here:

• \(i\) is the van 't Hoff factor which represents the number of particles the solute splits into.
• \(K_b\) is the ebullioscopic constant of the solvent.
• \(m\) is the molality of the solution.

The normal boiling point of the solvent is increased by the value of \(\Delta T_b\). This property relies directly on the number of solute molecules added, and it helps in determining molecular weights of unknown solutes by observing boiling point changes.

Freezing Point Depression

Freezing point depression occurs when the addition of a solute lowers the freezing point of the solvent. This is another colligative property that depends on the number of solute particles. The formula used to calculate the change in freezing point (\(\Delta T_f\)) is:

• \(\Delta T_f = i \cdot K_f \cdot m\)

Where:

• \(i\) is the van 't Hoff factor.
• \(K_f\) is the cryoscopic constant of the solvent.
• \(m\) is the molality of the solute.

This means that solutions will have a lower freezing point than the pure solvent. This principle is commonly used in applications like antifreeze in vehicles, where it helps prevent the liquid inside the engine from freezing in cold temperatures.

Molality

Molality (\(m\)) is a concentration term that's used in the study of colligative properties. It's defined as the number of moles of solute per kilogram of solvent:

• \(m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\)

Unlike molarity, which is dependent on the volume of the solution and can vary with temperature, molality remains constant because it refers to mass. This consistency makes molality particularly useful when dealing with problems involving temperature changes, like boiling point elevation or freezing point depression. Thus, understanding and calculating molality is crucial for accurately predicting how solutions will behave under different conditions.

Van 't Hoff Factor

The van 't Hoff factor (\(i\)) is an important concept in colligative properties, representing the number of particles into which a solute dissociates in a solution. For many non-electrolytes, \(i\) is typically 1 because they do not dissociate into smaller particles. However, electrolytes like sodium chloride (\(\text{NaCl}\)) split into ions, making their factor greater than one.
For example, NaCl dissociates into two ions: Na⁺ and Cl^-, giving it an \(i\) value of 2. This factor is crucial in calculations involving boiling point elevation or freezing point depression because it alters the number of particles considering the solute's dissociation characteristics. Understanding the van 't Hoff factor is essential for predicting how a solution's colligative properties will change based on the solute’s nature.