Problem 38 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Addition of \(0.643 \mathrm{~g}\) of a compound to \(50 \mathrm{~mL}\). of benzene (density : \(0.879 \mathrm{~g} / \mathrm{mL}\).) lowers the freezing point from \(5.51^{\circ} \mathrm{C}\) to \(5.03^{\circ} \mathrm{C}\). If \(K_{f}\) for benzene is \(5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\), calculate the molecular weight of the compound.

Step-by-Step Solution

Answer

The molecular weight of the compound is approximately 156.14 g/mol.

Step 1: Calculate Freezing Point Depression (ΔTf)

The freezing point depression, \( \Delta T_f \), can be calculated by subtracting the new freezing point from the normal freezing point of benzene.\[ \Delta T_f = 5.51^{\circ} \mathrm{C} - 5.03^{\circ} \mathrm{C} = 0.48^{\circ} \mathrm{C} \]

Step 2: Calculate Molality (m) Using ΔTf and Kf

The relationship between freezing point depression and molality is given by the formula \( \Delta T_f = K_f \times m \), where \( m \) is the molality. We rearrange the formula to solve for \( m \):\[ m = \frac{\Delta T_f}{K_f} = \frac{0.48^{\circ} \mathrm{C}}{5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}} \approx 0.09375 \mathrm{~mol} / \mathrm{kg} \]

Step 3: Calculate the Mass of Benzene in Kilograms

The mass of the benzene solvent is calculated by using its density. Given \( 50 \mathrm{~mL} \) of benzene and its density \( 0.879 \mathrm{~g/mL} \), the mass is:\[ \text{Mass of benzene} = 50 \mathrm{~mL} \times 0.879 \mathrm{~g/mL} = 43.95 \mathrm{~g} \]\[ = 0.04395 \mathrm{~kg} \]

Step 4: Calculate the Moles of Solute

Using the molality equation \( m = \frac{\text{moles of solute}}{\text{kg of solvent}} \), we solve for the moles of solute using the solvent's mass (in kg) from the previous step.\[ 0.09375 \mathrm{~mol} / \mathrm{kg} = \frac{\text{moles of solute}}{0.04395 \mathrm{~kg}} \]\[ \text{Moles of solute} = 0.09375 \times 0.04395 \approx 0.0041215 \mathrm{~mol} \]

Step 5: Calculate Molecular Weight of the Compound

The molecular weight (M) of the compound is calculated by using the mass of the solute and the number of moles computed in the previous step.\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{0.643 \mathrm{~g}}{0.0041215 \mathrm{~mol}} \approx 156.14 \mathrm{~g/mol} \]

Key Concepts

Molality

Molality is a key concept in understanding how solvents and solutes interact in solutions, specifically when it comes to freezing point depression. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which relies on the volume of the solution, molality is based on the mass of the solvent, making it particularly useful in scenarios where temperature changes are involved since it doesn't change with temperature variations.

To calculate molality, you need to know how much solute you have in moles and the mass of the solvent in kilograms. For example, in our problem, after determining the freezing point depression, we use the equation \[ ΔT_f = K_f \times m \] to find molality, where \( ΔT_f \) is the freezing point depression and \( K_f \) is the freezing point depression constant. Once we have these values, we rearrange the equation to solve for molality, \( m = \frac{ΔT_f}{K_f} \), giving us the concentration of solute in moles per kilogram of solvent. This is a crucial step for further calculations, such as determining the molecular weight of a solute.

Molecular Weight Calculation

Molecular weight, sometimes called molar mass, is the mass of a single molecule of a chemical compound. It is typically expressed in grams per mole (g/mol). The calculation of molecular weight is fundamental in chemistry for understanding how much of a substance is involved in a reaction or how much energy may be required to break a solution down.

In our problem, molecular weight calculation is derived from the relationship between weight, moles, and molality. First, you calculate the moles of the solute using the formula:\[ ext{moles of solute} = ext{molality} \times ext{mass of solvent in kg} \]After finding the number of moles, you can calculate the molecular weight using the formula:\[ ext{Molecular Weight} = \frac{ ext{Mass of solute in grams}}{ ext{Moles of solute}} \]This calculation helps identify the solute's size and how it may interact with solvents in various scenarios. For example, we found the molecular weight to be approximately \(156.14 \text{ g/mol}\) based on provided data. This understanding is vital for tasks such as quality control and developing chemical processes.

Colligative Properties

Colligative properties are characteristics of solutions that depend on the number of solute particles in a given amount of solvent, rather than the type of particles. These include key properties like freezing point depression, boiling point elevation, osmotic pressure, and vapor pressure lowering. Importantly, they depend on the concentration of solute particles, making them crucial for practical applications such as antifreeze formulation and food preservation.

Freezing point depression is a classic example of a colligative property. It's observed when a solute is added to a solvent, decreasing the solvent's freezing point. This is because solute particles disrupt the formation of the orderly solid structure needed to freeze the solvent. The formula that describes this phenomenon, discussed in the original exercise, is:\[ ΔT_f = K_f \times m \]where \( ΔT_f \) 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.

Overall, understanding colligative properties allows scientists and engineers to predict and manipulate the physical behaviors of solutions, which is essential in fields ranging from industrial applications to pharmaceuticals.