Problem 44 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution \(\mathrm{M}\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is \(0.9\) Given : Freezing point depression constant of water \(\left(K_{f}^{\text {water }}\right)\) $$ =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Freezing point depression constant of ethanol ( \(\left.K_{f}{\underline{\phantom{xx}}}^{\text {ethanol }}\right)\) $$ =2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Boiling point elevation constant of water \(\left(K_{b}^{\text {water }}\right)\) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Boiling point elevation constant of ethanol \(\left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Standard freezing point of water \(=273 \mathrm{~K}\) Standard freezing point of ethanol \(=155.7 \mathrm{~K}\) Standard boiling point of water \(=373 \mathrm{~K}\) Standard boiling point of ethanol \(=351.5 \mathrm{~K}\) Vapour pressure of pure water \(=32.8 \mathrm{~mm} \mathrm{Hg}\) Vapour pressure of pure ethanol \(=40 \mathrm{~mm} \mathrm{Hg}\) Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\) Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\) In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution \(\mathrm{M}\) is (a) \(39.3 \mathrm{~mm} \mathrm{Hg}\) (b) \(36.0 \mathrm{~mm} \mathrm{Hg}\) (c) \(29.5 \mathrm{~mm} \mathrm{Hg}\) (d) \(28.8 \mathrm{~mm} \mathrm{Hg}\)

Step-by-Step Solution

Answer

The vapour pressure of the solution M is approximately 39.3 mm Hg, so it's option (a).

Step 1: Identify Given Data

We are given the mole fraction of ethanol in the mixture, \( x_{\text{ethanol}} = 0.9 \). This implies that the mole fraction of water is \( x_{\text{water}} = 1 - x_{\text{ethanol}} = 0.1 \). We also know the vapour pressures of pure components: ethanol \( P_{0, \text{ethanol}} = 40\,\text{mm Hg} \) and water \( P_{0, \text{water}} = 32.8\,\text{mm Hg} \).

Step 2: Apply Raoult's Law

Raoult's Law states that the partial vapour pressure of each component in the solution is equal to the product of the mole fraction and the vapour pressure of the pure component. For ethanol, \( P_{\text{ethanol}} = x_{\text{ethanol}} \times P_{0, \text{ethanol}} = 0.9 \times 40 = 36\,\text{mm Hg} \). For water, \( P_{\text{water}} = x_{\text{water}} \times P_{0, \text{water}} = 0.1 \times 32.8 = 3.28\,\text{mm Hg} \).

Step 3: Calculate Total Vapour Pressure

The total vapour pressure of the solution \( M \) is the sum of the partial pressures of ethanol and water. Thus, \( P_{\text{solution}} = P_{\text{ethanol}} + P_{\text{water}} = 36 + 3.28 = 39.28\,\text{mm Hg} \).

Step 4: Choose the Closest Option

Compare our calculated total vapour pressure with the given options. The closest value to \( 39.28\,\text{mm Hg} \) is option (a) \( 39.3\,\text{mm Hg} \).

Key Concepts

Boiling Point Elevation

When a solute is dissolved in a solvent, it elevates the boiling point of the solution compared to the pure solvent. This phenomenon is called boiling point elevation. Boiling point elevation happens because the addition of solute causes a decrease in the solvent's vapor pressure, requiring more heat input to reach boiling.

In mathematical terms, the change in boiling point is calculated using the formula: \[\Delta T_b = i \cdot K_b \cdot m\]Where:\( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant of the solvent, and \( m \) is the molal concentration of the solution.

• Ebullioscopic constant \( K_b \) varies for each solvent. For water, \( K_b = 0.52 \ \mathrm{K \ kg \ mol^{-1}} \)
• Adding a solute always increases the boiling point as long as the solute does not dissociate, maintaining non-volatility.

Not surprisingly, this property has practical applications such as antifreeze solutions that raise boiling points in automotive radiators.

Freezing Point Depression

Freezing point depression is a colligative property where the presence of a solute lowers the freezing point of a solution. It's the reason why salt is spread on icy roads to prevent ice from forming. When a solute is added, the solution's vapor pressure is affected, leading to a lower freezing temperature.

The change in freezing point can be modeled using the formula:\[\Delta T_f = i \cdot K_f \cdot m\]Where:\( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the solute's molality.

• Cryoscopic constant \( K_f \) depends on the solvent. In the original exercise, \( K_{f}^{\text{water}} = 1.86 \ \mathrm{K \ kg \ mol^{-1}} \) and \( K_{f}^{\text{ethanol}} = 2.0 \ \mathrm{K \ kg \ mol^{-1}} \)
• This property helps in understanding why pure solvents freeze and solutions do not freeze at the same temperature.

This phenomenon is very useful in creating mixtures that need to remain liquid at even lower temperatures.

Raoult's Law

Raoult's Law is an essential principle in understanding colligative properties, specifically in determining the vapor pressures of solutions. It states that the partial vapor pressure of a component in a mixture is proportional to its mole fraction in the solution and the vapor pressure of the pure component.

The formula for Raoult's Law is:\[P_i = x_i \cdot P_{0,i} \]Where:\( P_i \) is the partial vapor pressure, \( x_i \) is the mole fraction, and \( P_{0,i} \) is the vapor pressure of the pure component.

• Raoult's Law applies well to ideal solutions where solute-solvent interactions do not significantly differ from solvent-solvent interactions.
• By using this law, it is possible to calculate the total vapor pressure of a solution, which is the sum of partial pressures of its components.
• This principle helps explain why solutions behave differently from their pure solvents and aids in calculating how components like ethanol and water mix.

Understanding Raoult’s Law is crucial for predicting how solutions will evaporate and boil, impacting both industrial and everyday applications.