Problem 47 - A Complete Resource Book in Chemistry for JEE Main

Question

By dissolving \(5 \mathrm{~g}\) substance in \(50 \mathrm{~g}\) of water, the decrease in freezing point is \(1.2^{\circ} \mathrm{C}\). The molal depression constant is \(1.85^{\circ} \mathrm{kg} \mathrm{mol}^{-1} .\) The molecular weight of substance is (a) \(105.4\) (b) \(118.2\) (c) \(137.2\) (d) \(154.2\)

Step-by-Step Solution

Step 1: Understand the Given Values

We are given:1. Mass of the solute (substance) = \(5 \text{ g}\).2. Mass of solvent (water) = \(50 \text{ g} = 0.05 \text{ kg}\).3. Depression in freezing point, \(\Delta T_f = 1.2 ^\circ \text{C}\).4. Molal depression constant, \(K_f = 1.85 ^\circ \text{kg} \text{mol}^{-1}\).

Key Concepts

Molal Depression Constant

The Molal Depression Constant, often represented as \(K_f\), is a crucial factor in calculating how the freezing point of a solvent is lowered when a solute is added. This constant is unique to each solvent. In simple terms, \(K_f\) helps determine how much the freezing point will drop per molal concentration of a solute present.

For instance, water has a \(K_f\) value of \(1.86\, ^\circ \text{C}\, \text{kg}\, \text{mol}^{-1}\), which means that for every mole of substance dissolved in 1 kg of water, the freezing point will decrease by \(1.86\, ^\circ \text{C}\). The value you are using depends on the nature of both the solute and the solvent. This constant helps us compare the effect of different solutes on the freezing point of a specific solvent.

Understanding \(K_f\) becomes particularly useful when we need to predict how a solution will behave under cold conditions or when trying to identify a solute based on its effect on the solvent's freezing point.

Depression in Freezing Point

The Depression in Freezing Point, or \(\Delta T_f\), is the change in the temperature at which a liquid freezes once a solute is added. When a solute dissolves in a solvent, it disrupts the solvent's natural ability to form a solid structure. As a result, the solvent's freezing point lowers.

This change is calculated using the formula: \[\Delta T_f = i \cdot K_f \cdot m\] where:

• \(i\) is the van 't Hoff factor (equal to 1 for non-electrolytes).
• \(K_f\) is the molal depression constant.
• \(m\) (molality) is the moles of solute per kilogram of solvent.

This formula shows that the more solute is present, the greater the decrease in the freezing point. Like wearing layers in the cold lowers your body's exposure to the temperature, adding more solute molecules 'shields' the solvent molecules from reaching their freezing point.

Knowing \(\Delta T_f\) is crucial for various real-world applications, such as preventing roads from freezing or designing antifreeze for cars.

Molecular Weight Calculation

Calculating the molecular weight of a substance through the freezing point depression is based on relationships involving the amount of solute and how it affects the freezing point. The formula that ties everything together is:\[M = \frac{K_f \cdot w}{\Delta T_f \cdot W}\] where:

• \(M\) is the molecular weight of the solute.
• \(K_f\) is the molal depression constant.
• \(w\) is the mass of solute (in grams).
• \(\Delta T_f\) is the observed freezing point depression (in °C).
• \(W\) is the mass of the solvent (in kilograms).

By plugging in the values, you compute \(M\), giving the molecular weight of the unknown substance. In practice, measuring \(\Delta T_f\) can help identify unknown solutes based on these principles.

This method is a handy tool in chemistry to find unknown substances by using their effects on the freezing point of a known solvent.