Problem 43 - 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 freezing point of the solution \(\mathrm{M}\) is (a) \(268.7 \mathrm{~K}\) (b) \(268.5 \mathrm{~K}\) (c) \(234.2 \mathrm{~K}\) (d) \(150.9 \mathrm{~K}\)

Step-by-Step Solution

Answer

The freezing point of the solution is 263.7 K (not among provided options).

Step 1: Understand the Problem

We need to find the freezing point of a solution M, which is a mixture of ethanol and water. The mole fraction of ethanol is 0.9, so the mole fraction of water is 0.1. We will be using the freezing point depression formula for this calculation.

Step 2: Identify Relevant Constants

Given that the freezing point depression constant for water, \(K_f^{\text{water}}\), is \(1.86 \text{ K kg mol}^{-1}\). Standard freezing point of water is \(273 \text{ K}\).

Step 3: Calculate Molality

Molality \(m\) is calculated as \(m = \frac{n_{\text{solute}}}{m_{\text{solvent}}}\). Here, ethanol acts as the solute and water as the solvent. Assume 1 mole of total solution: this means 0.9 moles of ethanol (solute) and 0.1 moles of water (solvent). Molality \(m = \frac{0.9}{(0.1 \times 18)} \text{ mol/kg} = 5 \text{ mol/kg}\).

Step 4: Apply Freezing Point Depression Formula

The formula for freezing point depression is \(\Delta T_f = K_f \times m\), where \(\Delta T_f\) is the decrease in freezing point, \(K_f\) is the freezing point depression constant, and \(m\) is the molality. Substitute the known values: \(\Delta T_f = 1.86 \text{ K kg mol}^{-1} \times 5 \text{ mol/kg} = 9.3 \text{ K}\).

Step 5: Calculate New Freezing Point

Subtract the \(\Delta T_f\) from the standard freezing point of water to get the new freezing point: \(273 \text{ K} - 9.3 \text{ K} = 263.7 \text{ K}\). Round as needed for final answer choices.

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property that occurs when a solute is dissolved into a solvent, leading to a decrease in the freezing point of the solution compared to the pure solvent. This happens because the presence of solute particles disrupts the formation of the solid structure of the solvent, thereby requiring a lower temperature to freeze. This principle is used in everyday applications, such as adding salt to roads to melt ice.

To calculate the depression in freezing point (\( \Delta T_f \)), you can use the formula:\[\Delta T_f = K_f \times m\]where:

• \( \Delta T_f \): the change in freezing point
• \( K_f \): the freezing point depression constant
• \( m \): the molality of the solution

The solution's new freezing point is determined by subtracting \( \Delta T_f \) from the pure solvent's freezing point. For a mixture of ethanol and water, you'll find that the significant presence of ethanol decreases the solution's freezing point significantly drawing from the calculations in the example.

Boiling Point Elevation

Boiling point elevation is another type of colligative property that describes the effect of solute particles on increasing the boiling point of a solvent. In a solution, the addition of non-volatile solute particles elevates the boiling point because more energy is needed for the solvent molecules to escape into the gas phase. This is why antifreeze is added to car radiators, preventing the coolant from boiling over in high temperatures.

The boiling point elevation can be calculated by the formula:\[\Delta T_b = K_b \times m\]where:

• \( \Delta T_b \): the change in boiling point
• \( K_b \): the boiling point elevation constant
• \( m \): the molality of the solution

In our example of a mixture of ethanol and water, you would need to use the specific \( K_b \) for water combined with the molality to determine the change. The resultant boiling point is then found by adding \( \Delta T_b \) to the pure solvent's boiling point.

Vapour Pressure

Vapour pressure is a key property of liquids that describes the pressure exerted by a vapor in thermodynamic equilibrium with its liquid at a given temperature. When solute particles are added to a solvent, the vapour pressure decreases because fewer solvent molecules are on the surface to evaporate. This phenomenon is responsible for both freezing point depression and boiling point elevation.

For mixtures like the ethanol-water solution, Raoult's Law helps predict the new vapour pressure:\[p = x_1p_1^0 + x_2p_2^0\]where:

• \( p \): the vapour pressure of the solution
• \( x_1, x_2 \): mole fractions of the components
• \( p_1^0, p_2^0 \): vapour pressures of the pure components

Understanding vapour pressure changes offers insight into why solutions behave differently from pure solvents, affecting everything from the physical properties to chemical reactivity.

Mole Fraction

The mole fraction is a way of expressing the composition of solutions. It is defined as the ratio of the number of moles of one component to the total number of moles of all components in the solution. The mole fraction is dimensionless and provides critical insights into how concentrated or dilute a particular component is.

For a solution, the mole fraction formula is:\[x_i = \frac{n_i}{n_{total}}\]where:

• \( x_i \): mole fraction of component i
• \( n_i \): moles of component i
• \( n_{total} \): total moles of all components

In the case of the ethanol-water mixture, with a mole fraction of ethanol at 0.9, it means ethanol is the major component affecting the solution's properties like freezing point and vapour pressure. This simple ratio becomes foundational in predicting and understanding colligative properties and solution behaviors.