Problem 33 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

On dissolving \(0.5 \mathrm{~g}\) of a non-volatile non-ionic solute to \(39 \mathrm{~g}\) of benzene, its vapour pressure decreases from \(650 \mathrm{~mm} \mathrm{Hg}\) to \(640 \mathrm{~mm}\) \(\mathrm{Hg}\). The depression of freezing point of Benzene (in \(\mathrm{K}\) ) upon addition of the solute is (Given data : Molar mass and the molal freezing point depression constant of benzene are \(78 \quad \mathrm{~g} \quad \mathrm{~mol}^{-1} \quad\) and \(5.12 \quad \mathrm{~K} \quad \mathrm{~kg} \quad \mathrm{~mol}^{-1}\), respectively)

Step-by-Step Solution

Answer

The depression of freezing point is approximately 1.009 K.

Step 1: Calculate the mole fraction of solute

Using Raoult's law, the decrease in vapor pressure is given by \( \Delta P = P^0 - P = 650 \mathrm{~mm} \mathrm{Hg} - 640 \mathrm{~mm} \mathrm{Hg} = 10 \mathrm{~mm} \mathrm{Hg} \). The mole fraction of solute \( x_2 \) is calculated as \( x_2 = \frac{\Delta P}{P^0} = \frac{10}{650} \approx 0.015385 \).

Step 2: Calculate the molality of the solution

The purity of solvent allows us to assume it doesn't participate in vapor pressure change: \( x_1 + x_2 = 1 \). Thus \( x_1 = 1 - x_2 = 0.984615 \). Knowing \( x_2 \) and the given molar mass of benzene (78 g/mol), calculate the amount (in moles) of solute: \( n_2 = \frac{\text{mass of solute}}{\text{molar mass}} = \frac{0.5}{m} \). Therefore, \( x_2 = \frac{n_2}{n_1 + n_2} \), using the approximation since solvent is present in excess: \( x_2 \approx \frac{n_2}{n_1} \rightarrow n_2 = x_2 \times n_1 \).

Step 3: Calculate molality (m)

Let \( m \) be the molar mass of solute. As determined previously, \( n_2 = 0.015385 \times 0.5 \approx 0.0076925 \) moles in benzene. Molality \( m = \frac{n_2}{\text{mass of solvent (in kg)}} = \frac{0.0076925}{0.039} \approx 0.197 \mathrm{~mol/kg} \).

Step 4: Calculate depression in freezing point

Finally, use the formula for depression in freezing point: \( \Delta T_f = K_f \times m \), where \( K_f = 5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \). Thus \( \Delta T_f = 5.12 \times 0.197 \approx 1.00864 \mathrm{~K} \).

Key Concepts

Raoult's Law

Raoult's Law is pivotal when learning about solutions and their properties. It describes how the vapor pressure of a solvent changes when a solute is added. In essence, it states that the partial vapor pressure of each component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

This highlights that the presence of a solute lowers the overall vapor pressure of the solvent. This is because some of the surface area that was originally occupied by the solvent molecules is now taken up by solute particles.

Therefore, the new vapor pressure of the solution is slightly lower than the solvent's original vapor pressure. This principle is vital in calculating changes such as the depression of the freezing point.

Mole Fraction

Mole fraction is one way to express the composition of a solution. It is used extensively in the context of Raoult's Law. The mole fraction is defined as the ratio of the number of moles of a component to the total number of moles in the solution.

This number doesn't have units, which makes it convenient to use in various calculations. For example, in the exercise, the mole fraction of the solute was used to determine the change in vapor pressure according to Raoult's Law. By knowing the mole fractions, we ensure an accurate construction of the chemical balance in the solution.

It's important to remember that the sum of the mole fractions of all components in a solution must be equal to 1.

Molality

Molality ( extit{m}) is another unit to express concentration, and it plays a significant role in colligative properties like freezing point depression. It is defined as the number of moles of solute per kilogram of solvent.

Molality is advantageous because it doesn't change with temperature, unlike molarity, which involves volume that can expand or contract with temperature changes. This property makes it ideal for use in calculations involving changes such as the depression of freezing point.

In our given exercise, molality was calculated by dividing the moles of solute by the mass of the solvent in kilograms. It served as a key step in determining how the freezing point of benzene was affected.

Vapor Pressure

Vapor pressure is a measure of the tendency of a substance to evaporate. In solutions, the vapor pressure is influenced by the presence of solute particles according to Raoult's Law.

The vapor pressure of a solution is typically lower than that of the pure solvent, due to the solute particles hindering the escape of solvent molecules from the liquid into the gaseous phase.

In the exercise, vapor pressure measurement was important to understand how much the vapor pressure decreased upon the addition of a solute. The calculation started with an initial vapor pressure, and after the solute was added, the reduced vapor pressure was observed. This change is essential as it feeds directly into other property changes such as boiling point elevation and freezing point depression.