Problem 194 - A Complete Resource Book in Chemistry for JEE Main

Question

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration \(0.1 \mathrm{M}\) is \(100 \Omega\). The conductivity of this solution is \(1.29 \mathrm{~S} \mathrm{~m}^{-1}\). Resistance of the same cell when filled with \(0.2 \mathrm{M}\) of the same solution is \(520 \Omega\). The molar conductivity of \(0.02 \mathrm{M}\) solution of the electrolyte will be (a) \(124 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) (b) \(1240 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) (c) \(1.24 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) (d) \(12.4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The correct answer is (b) \( 1240 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \).

Step 1: Calculate cell constant

The cell constant \( K \) can be found using the formula \( K = R_1 \times \kappa_1 \), where \( R_1 = 100 \ \Omega \) is the resistance of the 0.1 M solution and \( \kappa_1 = 1.29 \ \text{S m}^{-1} \) is the conductivity. Thus, \( K = 100 \times 1.29 = 129 \ \text{m}^{-1} \).

Step 2: Find conductivity for 0.2 M solution

We can determine the conductivity \( \kappa_2 \) of the 0.2 M solution using the cell constant \( K \) and the measured resistance \( R_2 = 520 \ \Omega \). By using the formula \( \kappa_2 = \frac{K}{R_2} \), we find \( \kappa_2 = \frac{129}{520} = 0.2481 \ \text{S m}^{-1} \).

Step 3: Compute molar conductivity for 0.02 M solution

The molar conductivity \( \Lambda_m \) can be calculated using the formula \( \Lambda_m = \frac{\kappa}{c} \), where \( \kappa = 0.2481 \ \text{S m}^{-1} \) and \( c = 0.02 \ \text{mol L}^{-1} = 0.02 \ \text{mol m}^{-3} \). Thus, \( \Lambda_m = \frac{0.2481}{0.02} = 12.405 \ \text{S m}^2 \text{mol}^{-1} \).

Step 4: Convert to given units and compare options

The calculated molar conductivity is \( 12.405 \times 10^{-2} \ \text{S m}^2 \text{mol}^{-1} \). Converting this into \( \text{S } \text{m}^2 \text{mol}^{-1} \times 10^{-4} \), it becomes \( 1240 \times 10^{-4} \ \text{S m}^2 \text{mol}^{-1} \) by multiplying \( 12.405 \) by \( 100 \). This matches option (b).

Key Concepts

Conductivity

Conductivity is a measure of a material's ability to conduct an electric current. In the context of solutions, it refers to the extent to which an electrolyte solution can conduct electricity. Conductivity is represented by the symbol \( \kappa \) and is measured in Siemens per meter (S/m).
Conductivity provides insights into the nature of ions in solution and their concentration. The higher the number of ions, the higher the conductivity of the solution. For instance:

• High ion concentration increases conductivity.
• Strong electrolytes fully dissociate into ions, showcasing higher conductance.
• Conductivity changes with concentration changes in the solution.

Understanding conductivity is essential for solving various chemistry problems, including those in JEE Main Chemistry exams.

Resistance

Resistance is a fundamental concept in both physics and chemistry that refers to the opposition a material offers to the flow of electric current. It is measured in ohms (\( \Omega \)). In the context of a conductivity cell, resistance is determined by how difficult it is for the current to pass through the solution.
The relationship between resistance, cell constant, and conductivity is crucial:

• Resistance \( R \) in a solution impacts the current flow and is inversely related to conductivity \( \kappa \).
• Higher resistance surfaces suggest fewer ions or weaker electrolyte conditions.
• Resistance's formula in a cell context relates to \( R = \frac{K}{\kappa} \), where \( K \) is the cell constant.

Grasping the concept of resistance helps students analyze and solve problems concerning electrolyte solutions.

Cell Constant

The cell constant, symbolized as \( K \), is a crucial parameter in electrochemistry. It provides a measure of a cell's geometry, specifically its dimensions related to electrode spacing and surface area. The unit of the cell constant is per meter (\( \text{m}^{-1} \)).
The formula to calculate the cell constant is \( K = R \times \kappa \).
Here's why the cell constant is vital:

• The cell constant relates directly to how the resistance and conductivity of a cell's solution can be measured.
• If the cell dimensions change, the constant adjusts, affecting calculations of conductivity.
• It ensures precise measurements in different calibrated cells used during experiments.

Practically applying the cell constant enables accurate readings and results in electrolyte solution assessments.

Electrolyte Solution

An electrolyte solution is one that conducts electricity due to the presence of free-moving charged particles known as ions. These ions result from the dissociation of compounds, such as salts, acids, or bases, in a solvent, like water.
Electrolytes play a significant role in chemistry:

• Strong electrolytes dissociate completely, leading to better conductivity.
• Weak electrolytes partially dissociate, offering lesser conductivity.
• The concentration of the solution directly affects its conductivity.

Understanding electrolyte solutions is key for JEE Main Chemistry, as topics often involve calculating conductivity, resistance, and molar conductivity using solutions at various concentrations.

JEE Main Chemistry

JEE Main Chemistry is a national level exam in India designed to test students' understanding and application of core chemical concepts. It covers a range of topics, including the physical, organic, and inorganic branches.
Key points for success in JEE Main Chemistry:

• Grasp fundamental principles, such as those involving conductivity and solutions.
• Practice problems regularly to become familiar with question patterns.
• Focus on understanding concepts instead of rote memorization.

Mastering these areas not only prepares students well for the exam but also builds a robust conceptual foundation in chemistry.