Problem 123 - A Complete Resource Book in Chemistry for JEE Main

Question

The freezing point of a solution prepared by dissolving \(20.5 \mathrm{~g}\) of a non- volatile non- electrolyte with empricial formula \(\left(\mathrm{C}_{3} \mathrm{H}_{2}\right)\) in \(400 \mathrm{~g}\) of benzene is \(4.33^{\circ} \mathrm{C}\). The benzene used to prepare the solution freezes at \(5.48^{\circ} \mathrm{C}\), using the same thermometer. (Given Kf of benzene \(=5.1 \mathrm{~kg} \mathrm{~K} \mathrm{~mol}^{-1}, \mathrm{~Kb}\) of benzene \(=2.53\) \(\mathrm{kg} \mathrm{K} \mathrm{mol}^{-1}\) and boiling point of benzene \(\left.80.2^{\circ} \mathrm{C}\right)\). (a) The boiling point of the solution will be \(80.17^{\circ} \mathrm{C}\). (b) The boiling point of the solution will be \(80.77^{\circ} \mathrm{C}\) (c) The molecular formula of non electrolyte is \(\mathrm{C}_{18} \mathrm{H}_{12} .\) (d) The molecular formula of non- electrolyte is \(\mathrm{C}_{15} \mathrm{H}_{12}\)

Step-by-Step Solution

Answer

(b) The boiling point is 80.77°C. (c) The molecular formula is C18H12.

Step 1: Calculate Freezing Point Depression

The freezing point depression (              \( \Delta T_f \)) can be calculated using the formula: \( \Delta T_f = K_f \cdot m \), where \( K_f = 5.1 \ \text{kg K mol}^{-1} \) is the cryoscopic constant and \( m \) is the molality. Given that the initial freezing point of benzene is \(5.48^{\circ} \text{C}\) and the freezing point of the solution is \(4.33^{\circ} \text{C}\), the freezing point depression is \( \Delta T_f = 5.48 - 4.33 = 1.15^{\circ} \text{C}\).

Step 2: Calculate Molality of the Solution

The relation \( \Delta T_f = K_f \cdot m \) helps to find molality, so solve for molality \( m = \frac{\Delta T_f}{K_f} \). With \( \Delta T_f = 1.15 \ \text{C} \) and \( K_f = 5.1 \ \text{kg K mol}^{-1} \), we calculate: \[ m = \frac{1.15}{5.1} = 0.2255 \ \text{mol/kg} \].

Step 3: Determine Moles of Solute

Calculate the moles of solute using: \( \text{moles} = m \times \text{mass of solvent in kg} \). With \( m = 0.2255 \ \text{mol/kg} \) and \( \text{mass of benzene} = 400 \ \text{g} = 0.4 \ \text{kg} \), the moles of solute are \[ 0.2255 \times 0.4 = 0.0902 \ \text{moles} \].

Step 4: Calculate Molar Mass of Solute

Find the molar mass using the formula: \( \text{molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} \). Given \( \text{mass of solute} = 20.5 \ \text{g} \) and \( \text{moles of solute} = 0.0902 \ \text{moles} \), \[ \text{molar mass} = \frac{20.5}{0.0902} \approx 227.16 \ \text{g/mol} \].

Step 5: Determine Molecular Formula

The empirical formula mass of \( \text{C}_3\text{H}_2 \) is \( 3 \times 12 + 2 \times 1 = 38 \ \text{g/mol} \). Divide the molar mass of the compound by the empirical formula mass to find the number of empirical formula units, \( n = \frac{227.16}{38} \approx 6 \). Hence, the molecular formula is \( (\text{C}_3\text{H}_2)_6 = \text{C}_{18}\text{H}_{12} \).

Step 6: Calculate Boiling Point Elevation

Use the equation \( \Delta T_b = K_b \cdot m \) to find the boiling point elevation. With \( K_b = 2.53 \ \text{kg K/mol} \) and \( m = 0.2255 \ \text{mol/kg} \), \[ \Delta T_b = 2.53 \times 0.2255 \approx 0.57 \ \text{C} \].

Step 7: Calculate Boiling Point of Solution

The boiling point of the pure benzene is \(80.2^{\circ} \text{C}\). Add the boiling point elevation to this to get the boiling point of the solution: \( 80.2 + 0.57 = 80.77^{\circ} \text{C} \). Thus, the solution's boiling point matches the choice (b).

Key Concepts

Boiling Point Elevation

Boiling Point Elevation is a fascinating colligative property that occurs when a non-volatile solute is added to a solvent, such as benzene. This addition raises the boiling point of the liquid. When we dissolve a solute in a solvent, the vapor pressure decreases, requiring more heat to reach the boiling point, thus increasing the temperature.
The formula to calculate the boiling point elevation is \[ \Delta T_b = K_b \cdot m \], where

• \( \Delta T_b \) is the boiling point elevation.
• \( K_b \) is the ebullioscopic constant of the solvent, with units \( \text{kg} \, \text{K/mol} \).
• \( m \) is the molality of the solution.

In the presented problem, the boiling point elevation calculation results in a value of \(0.57^{\circ} \text{C}\). This value is then added to the pure benzene's boiling point \(80.2^{\circ} \text{C}\), providing us with the solution's boiling point of \(80.77^{\circ} \text{C}\). This illustrates how solutes impact the boiling characteristics of a solvent.

Molality

Molality is a key concept in understanding colligative properties, like freezing point depression and boiling point elevation. It measures the concentration of a solute in a solution. Differing from molarity, molality is not impacted by temperature changes as it depends on the mass of the solvent, not its volume.
Molality is defined as:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kilograms}} \]. This ensures accurate and consistent measurements in varying thermal conditions.
In the provided example, we initially found the freezing point depression, \(\Delta T_f\), and used the formula: \[ m = \frac{\Delta T_f}{K_f} \]. Substituting in the known values of \(\Delta T_f = 1.15^{\circ} \text{C}\) and \(K_f = 5.1 \text{kg K/mol}\), we calculated a molality of \(0.2255 \text{ mol/kg}\). This value is key to determining both freezing and boiling points in solutions.

Molar Mass Calculation

Molar mass calculation is vital to identifying unknown compounds and involves determining the molar mass from the mass and moles of a substance. It provides the mass of one mole of a given substance and plays a crucial role in stoichiometry and reaction predictions.
The formula used is: \[ \text{Molar Mass} = \frac{\text{mass of solute}}{\text{moles of solute}} \]. This gives an essential insight into the composition of substances.
Based on the information in the exercise, we know the mass of the solute as \(20.5 \text{g}\) and calculated the moles earlier as \(0.0902 \text{mol}\). Therefore, the molar mass is \[ \frac{20.5 \text{g}}{0.0902 \text{mol}} = 227.16 \text{g/mol} \]. Accurate molar mass calculation not only aids in determining empirical and molecular formulas but also supports understanding the physical and chemical behavior of compounds.

Molecular Formula Determination

Molecular formula determination involves deriving the actual number of each type of atom present in a molecule from its empirical formula and molar mass.
Firstly, we need the empirical formula's molar mass, which is calculated by adding the atomic masses of its constituent elements. In our case, for \(\text{C}_3\text{H}_2\), the molar mass is \(3 \times 12 + 2 \times 1 = 38 \text{g/mol}\).
Using the molar mass obtained, we determine the molecular formula by: \[ n = \frac{\text{molar mass of compound}}{\text{molar mass of empirical formula}} \]. That is, \[ n = \frac{227.16}{38} \approx 6 \]. Therefore, our molecular formula becomes \((\text{C}_3\text{H}_2)_6\), simplifying to \(\text{C}_{18}\text{H}_{12}\).
This process provides a deeper insight into the compound's structure, revealing essential information for further chemical analysis and applications.