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) \(12.4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)

Short answer

The molar conductivity of the 0.02 M solution is 1240 x 10^-4 S m^2 mol^-1 (option b).

Step-by-step solution

Understanding the Problem

We are given two different concentrations of an electrolyte solution and their corresponding resistances in a conductivity cell. We also know the conductivity of one solution and need to find the molar conductivity of another solution with a different concentration.

Calculating the Cell Constant

The cell constant (G*) can be determined using the formula: \( G^* = ext{conductance} \times R \), where R is resistance. From the given data, for the 0.1 M solution with known conductivity, the cell constant can be calculated as: \( G^* = rac{1}{100 \, \Omega} \times 1.29 \, \mathrm{S} \, \mathrm{m}^{-1} \).

Determining Conductance

Conductance (C) is the reciprocal of resistance (R). Therefore, the conductance of the 0.2 M solution is \( C = \frac{1}{520 \, \Omega} \).

Finding Conductivity of 0.2 M Solution

Since we know the cell constant, the conductivity (κ) of the 0.2 M solution is \( κ = C \times G^* \). Use the conductance found in the previous step and the previously calculated cell constant to compute this conductivity.

Calculating Molar Conductivity

Molar conductivity (Λ) is given by the formula: \( Λ = \frac{κ}{c} \), where c is the concentration. Use the calculated conductivity of the 0.02 M solution and plug it into this formula to obtain the molar conductivity.

Verify and Select Correct Option

Compare the calculated molar conductivity with the options provided: (a) \(124 \times 10^{-4} \)(b) \(1240 \times 10^{-4} \)(c) \(1.24 \times 10^{-4} \)(d) \(12.4 \times 10^{-4} \). Choose the option that matches the calculated value.

Key concepts

Electrolyte Solution

Electrolyte solutions are a key component in the study of chemistry, playing a crucial role in conducting electricity. An electrolyte is a compound that, when dissolved in a solvent, dissociates into ions. These ions are responsible for conducting electricity through the solution.
For example, common salts, acids, and bases are electrolytes. Their ability to conduct electric current makes them of significant interest in fields like electrochemistry. In the given exercise, electrolytes are in a solution, where their concentration can affect electrical resistance and conductance. Knowing how these factors interplay is vital for calculating properties like molar conductivity.

Conductivity Cell

A conductivity cell is a device used to measure the conductivity of an electrolyte solution. It comprises two electrodes and is submerged in the liquid whose conductance is to be measured.
During the process, a voltage is applied across the electrodes, and the current flowing through the solution is measured. This current helps determine how well the ions in the solution can carry electric charge, also known as conductivity.
The reliability of a conductivity cell is crucial, as it provides the necessary data to determine properties like resistance and cell constant for further analysis.

Cell Constant

The cell constant is a crucial parameter that describes the geometric configuration of the conductivity cell. It is defined as the ratio of the distance between the electrodes to the surface area of the electrodes.
Mathematically, it is expressed as: \[ G^* = \frac{1}{R} \times \kappa \]
where \( R \) is the resistance and \( \kappa \) is the conductivity of the solution.
In practice, the cell constant varies with changes in the design of the cell, making it an essential factor in calculating the exact conductivity of a solution. The cell constant is needed to convert the measured conductance into conductivity.

Resistance

Resistance is a measure of how much a material opposes the flow of electric current. In the context of electrolyte solutions, resistance is mainly determined by the concentration of ions available to carry a current.
An electrolyte's resistance is measured in ohms (\( \Omega \)) and is inversely related to conductance. Higher resistance typically means fewer ions are available to facilitate the flow of current.
In our example, comparing resistances at different concentrations helps us understand how concentration affects the solution's ability to conduct electricity.

Conductance

Conductance is essentially the flip side of resistance. It measures how easily electricity can pass through a material. Conductance is measured in siemens (\( \mathrm{S} \)), and is the reciprocal of resistance.
This concept is crucial when working with electrolyte solutions as it helps us quantify how different concentrations can influence the flow of electric current.
For a solution with a lower resistance, the conductance will be higher, indicating a better ability to carry electrical current. By understanding conductance, we can better manipulate and predict how solutions will behave under different conditions.

Concentration

Concentration refers to the amount of solute present in a solution, usually expressed in molarity (M), which is moles per liter. In electrolyte solutions, concentration is vital because it directly impacts the number of ions available to conduct electricity.
A higher concentration means more ions, which typically increases the solution's conductance. However, it's essential to remember that beyond a certain point, adding more solute may not linearly increase conductance due to inter-ionic interactions.
Understanding concentration allows chemists to tailor a solution's electrical properties for specific applications, ensuring optimal performance in conductivity experiments.

JEE Main Chemistry

Chemistry, as part of the Joint Entrance Examination (JEE) Main, involves various fundamental concepts, including electrolyte solutions and their properties. This entrance test is a critical assessment for students aiming to enter engineering programs in India.
The examination covers a vast syllabus, demanding a solid understanding of principles such as molar conductivity, conductance, and chemical equilibria.
For students, mastering these concepts not only helps in solving problems like those discussed here but also enhances their analytical skills, preparing them for advanced studies and technical challenges in engineering.