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 \(\mathrm{Kf}\) 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 \(80.2^{\circ} \mathrm{C}\) ). (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, and (c) the molecular formula is C₁₈H₁₂.

Step 1: Calculate Molality

The freezing point depression formula is \[\Delta T_f = K_f \cdot m\]where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the freezing point depression constant of the solvent, and \(m\) is the molality of the solution.First, find \(\Delta T_f\):\[\Delta T_f = 5.48^{\circ}C - 4.33^{\circ}C = 1.15^{\circ}C\]Solving for molality \(m\):\[m = \frac{\Delta T_f}{K_f} = \frac{1.15}{5.1} = 0.225 \text{ mol/kg}\]

Step 2: Determine Moles of Solute

The molality formula \(m\) is \[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\]Rearranging to find moles of solute:\[\text{moles of solute} = m \times \text{mass of solvent in kg} = 0.225 \times 0.400 = 0.09 \text{ moles}\]

Step 3: Calculate Molar Mass of Solute

The number of moles is related to mass and molar mass (M) by the equation:\[\text{moles} = \frac{\text{mass}}{M}\]Solving for M:\[M = \frac{\text{mass}}{\text{moles}} = \frac{20.5}{0.09} = 227.78 \text{ g/mol}\]

Step 4: Find Molecular Formula of Non-Electrolyte

The empirical formula \(\text{C}_3\text{H}_2\) has a molar mass of:\[(3 \times 12.01) + (2 \times 1.01) = 38.05 \text{ g/mol}\]The molecular formula is a multiple of the empirical formula.\[x \times 38.05 = 227.78\]Solve for \(x\):\[x = \frac{227.78}{38.05} \approx 6\]Thus, the molecular formula is:\[\text{C}_{3 \times 6}\text{H}_{2 \times 6} = \text{C}_{18}\text{H}_{12}\]

Step 5: Calculate Boiling Point Elevation and Solution Boiling Point

The boiling point elevation formula is:\[\Delta T_b = K_b \cdot m\]Calculate \(\Delta T_b\):\[\Delta T_b = 2.53 \times 0.225 = 0.56925\]Calculate the new boiling point:\[T_b = 80.2 + 0.56925 = 80.76925 \approx 80.77^{\circ}C\]

Key Concepts

Freezing Point Depression

When you dissolve a solute in a solvent, it lowers the temperature at which the solvent will freeze. This phenomenon is known as freezing point depression, a colligative property, meaning it depends on the number of solute particles in a given amount of solvent, not the type of particles. This happens because the presence of solute particles disrupts the formation of the solid structure of the solvent molecules as they freeze.

To calculate the change in freezing point, the formula used is \[\Delta T_f = K_f \cdot m\] - \(\Delta T_f\) represents the change in freezing point. - \(K_f\) is the freezing point depression constant, specific to each solvent. - \(m\) is the molality of the solution.

In the problem, the solution's freezing point was lowered from benzene's natural freezing point of \(5.48^{\circ}C\) to \(4.33^{\circ}C\). The calculated \(\Delta T_f\) was \(1.15^{\circ}C\), allowing us to solve for the molality, an essential step in further calculating the properties of the solute.

Boiling Point Elevation

Boiling point elevation is another colligative property, akin to freezing point depression, influenced by the number of particles in a solution. When a solute is added to a solvent, the boiling point of the solvent increases. This elevation occurs because additional energy (heat) is needed to overcome the vapor pressure decrease caused by the solute particles.

To calculate the increase in boiling point, the formula used is \[\Delta T_b = K_b \cdot m\] - \(\Delta T_b\) is the increase in boiling point. - \(K_b\) is the boiling point elevation constant for the solvent. - \(m\) is the molality of the same solution.

In the exercise provided, the boiling point elevation was calculated based on benzene's original boiling point of \(80.2^{\circ}C\) using the molality and the \(K_b\) value. The final boiling point of the solution was thus adjusted to approximately \(80.77^{\circ}C\). This shows how even a small amount of solute can significantly change the physical properties of a solvent.

Molality

Understanding molality is crucial for solving problems related to colligative properties. Molality, symbolized as \(m\), is defined as the moles of solute per kilogram of solvent. One of its advantages over other concentration measures is that it remains constant with temperature because it depends on mass, not volume.

To find molality, use: \[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\] This equation helps in determining the extent of freezing point depression or boiling point elevation in a solution. In the context provided, the calculated molality of the benzene solution was found to be \(0.225 \text{ mol/kg}\). This value is then used to tailor the calculations for changes in the freezing and boiling points.

Empirical and Molecular Formulas

The distinction between empirical and molecular formulas is foundational in chemistry. The empirical formula represents the simplest whole-number ratio of elements in a compound. For example, \(\text{C}_3\text{H}_2\) is an empirical formula, indicating carbon and hydrogen in a 3:2 ratio.

The molecular formula is either the same as or a multiple of the empirical formula, showing the actual number of atoms in a molecule. It is derived by comparing the molar mass of the empirical formula with the experimentally determined molar mass.

In the problem, the molecular formula \(\text{C}_{18}\text{H}_{12}\) was found by calculating the multiple of the empirical formula's molar mass \(38.05 \text{ g/mol}\) that equals the observed molar mass \(227.78 \text{ g/mol}\). This gives a factor of approximately 6, thus multiplying the subscripts in the empirical formula by 6 to deduce the correct molecular formula for the non-electrolyte.