Problem 78 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(0.44 \mathrm{~g}\) of substance dissolved in \(22.2 \mathrm{~g}\) of benzene lowers the freezing point of benzene by \(0.567^{\circ} \mathrm{C}\), then the molecular mass of substance is, (the molal depression constant \(=5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\) ) (a) \(128.4\) (b) \(156.6\) (c) \(178.9\) (d) \(232.4\)

Step-by-Step Solution

Answer

The molecular mass of the substance is 178.9 g/mol, option (c).

Step 1: Convert grams into kilograms

To use the formula for freezing point depression, convert the mass of benzene into kilograms. Given is 22.2 g of benzene, which is equivalent to 0.0222 kg.

Step 2: Use the freezing point depression formula

The formula for freezing point depression is \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( i \) is the van 't Hoff factor (assumed to be 1 for non-electrolytes), \( K_f \) is the molal depression constant, and \( m \) is the molality. We need to find \( m \), so rearrange to \( m = \frac{\Delta T_f}{i \cdot K_f} \). Substituting the values, \( m = \frac{0.567}{5.12} = 0.1105 \text{ mol/kg} \).

Step 3: Calculate the number of moles of solute

The molality \( m \) is designated as moles of solute per kilogram of solvent. Rearrange \( m = \frac{n}{\text{kg of solvent}} \) to find \( n \), the number of moles: \( n = m \cdot \text{kg of solvent} = 0.1105 \text{ mol/kg} \cdot 0.0222 \text{ kg} = 0.002454 \text{ mol} \).

Step 4: Calculate the molecular mass of the substance

Molecular mass is determined by dividing the given mass of the substance by the number of moles: \( \text{molecular mass} = \frac{0.44 \text{ g}}{0.002454 \text{ mol}} \approx 179.3 \text{ g/mol} \). Round to the closest option from the given choices, which is (c) 178.9.

Key Concepts

Molecular Mass Calculation

Calculating molecular mass can seem tricky, but with a systematic approach, it becomes simple. In essence, the molecular mass is determined by dividing the mass of the solute (in grams) by the number of moles that the solute represents.

• First, identify the given mass of the substance you are examining. In our exercise, that's the substance dissolved in benzene, which is 0.44 g.
• Next, calculate the number of moles. You accomplish this by using the molality of the solution and the mass of the solvent. In the step-by-step solution, the number of moles was calculated as 0.002454 mol.
• Finally, compute the molecular mass: divide the substance's mass (0.44 g) by the moles (0.002454 mol), resulting in approximately 179.3 g/mol.

Notice that the calculated value is rounded to match one of the given answer options, which is 178.9 g/mol.

Molal Depression Constant

The molal depression constant, often denoted as \( K_f \), is a crucial factor when analyzing changes in freezing points due to the dissolution of substances.This constant represents the extent to which the freezing point of a solvent decreases for every mole of solute dissolved per kilogram of solvent.

• Each solvent has a unique \( K_f \) value. For benzene, as given in the problem, it's \( 5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg} \).
• In the context of freezing point depression, the \( K_f \) value allows us to calculate how much the freezing point has been lowered due to the dissolved substance.
• What's more, it's vital when using the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is typically 1 for non-electrolytes, as assumed here.

Understanding the molal depression constant is essential for accurately predicting freezing point changes and calculating the molecular mass in such exercises.

Molality in Chemistry

Molality is a concentration term that relates the amount of solute to the solvent's mass, used especially in studies involving temperature changes like freezing point depression.Defining molality (\( m \)) is straightforward: it's the moles of solute per kilogram of solvent. It differs from molarity, which is moles per liter of solution.

• In the exercise provided, the focus was on a solution where 0.44 g of a substance was dissolved in 22.2 g of benzene.
• After converting the solvent mass to kilograms (0.0222 kg), we find the molality using the formula rearranged as \( m = \frac{\Delta T_f}{i \cdot K_f} \).
• Given \( \Delta T_f = 0.567^{\circ} \mathrm{C} \) and \( K_f = 5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg} \), calculating \( m \) results in 0.1105 mol/kg. This number is then used to trace back to the solute's moles.

By concentrating on molality, we achieve a clear view of how much a solute affects its solvent, critical for solutions involving temperature changes.