Problem 204 - A Complete Resource Book in Chemistry for JEE Main

Question

Resistance of \(0.2 \mathrm{M}\) solution of an electrolyte is \(50 \Omega\) The specific conductance of the solution is \(1.4 \mathrm{~S} \mathrm{~m}^{-1}\). The resistance of \(0.5 \mathrm{M}\) solution of the same electrolyte is \(280 \Omega\) The molar conductivity of \(0.5 \mathrm{M}\) solution of the electrolyte in \(\mathrm{S} \mathrm{mt}^{2} \mathrm{~mol}^{-1}\) is: (a) \(5 \times 10^{3}\) (b) \(5 \times 10^{2}\) (c) \(5 \times 10^{-4}\) (d) \(5 \times 10^{-3}\)

Step-by-Step Solution

Answer

The molar conductivity of the 0.5 M solution is \(5 \times 10^{-1} \) or 0.5 S m² mol⁻¹.

Step 1: Understand the Given Values

We are given the resistance for two solutions:- For a 0.2 M solution, the resistance is 50 Ω and the specific conductance (kappa, \( \kappa \)) is \(1.4 \, \mathrm{S} \, \mathrm{m}^{-1}\).- For a 0.5 M solution, the resistance is 280 Ω.We need to find the molar conductivity (\( \Lambda_m \)) of the 0.5 M solution.

Step 2: Calculate the Conductance

Conductance (G) is the reciprocal of resistance (R). Therefore, for the 0.5 M solution we have:\[G = \frac{1}{R} = \frac{1}{280} \, \Omega^{-1}\]

Step 3: Use the Formula for Conductance and Specific Conductance

The relationship between conductance (G), specific conductance (\( \kappa \)), and cell constant (C) is given by:\[\kappa = G \cdot C\]For the 0.2 M solution, we have:\[1.4 \, \mathrm{S} \, \mathrm{m}^{-1} = \left(\frac{1}{50}\right) \cdot C\]Thus, the cell constant is:\[C = 1.4 \times 50 \]

Step 4: Calculate the Cell Constant

Compute the cell constant C:\[C = 1.4 \times 50 = 70 \, \mathrm{m}^{-1}\]

Step 5: Calculate the Specific Conductance for the 0.5 M Solution

Now, apply the cell constant to find the specific conductance for the 0.5 M solution. \[\kappa = G \cdot C = \left(\frac{1}{280}\right) \cdot 70\]

Step 6: Solve for Specific Conductance

Substitute values to find \(\kappa\):\[\kappa = \frac{70}{280} = 0.25 \, \mathrm{S} \, \mathrm{m}^{-1}\]

Step 7: Calculate Molar Conductivity

Molar conductivity (\(\Lambda_m\)) is related to specific conductance and concentration by the formula:\[\Lambda_m = \frac{\kappa}{c}\]where \(c = 0.5 \, \mathrm{M}\). Substitute the known values:\[\Lambda_m = \frac{0.25}{0.5} \]

Step 8: Solve for Molar Conductivity

Complete the calculation:\[\Lambda_m = \frac{0.25}{0.5} = 0.5 \, \mathrm{S} \, \mathrm{m}^2 \, \mathrm{mol}^{-1}\]

Key Concepts

Specific Conductance

Specific conductance, also known as conductivity, measures how well a solution can conduct electricity. It is represented by the Greek letter \( \kappa \) and usually expressed in units of Siemens per meter (\( \mathrm{S} \, \mathrm{m}^{-1} \)).
Specific conductance depends on the concentration and nature of the ions in the solution. Higher ion concentration generally means higher conductivity.

To understand specific conductance better, picture it like how efficiently current flows through a solution:

• More ions or charged particles in a solution increase conductivity.
• The more free ions, the easier it is for an electric current to pass.

Learning how to calculate specific conductance helps chemists understand the behavior of solutions. This concept is essential in finding other properties like molar conductivity, especially in solutions of electrolytes.

Electrolyte Resistance

Electrolyte resistance is the opposition that a solution presents to the flow of electrical current. Measured in Ohms (\( \Omega \)), it's a vital concept in electrochemistry, helping us gauge a solution's behavior.
This resistance arises due to the solution's inherent properties and concentration of ions:

• Greater ion concentration usually means lower resistance, as ions facilitate electricity flow.
• Larger resistance indicates that the solution is less conductive.

In the exercise, resistance values provide essential data to calculate other parameters like molar conductivity. Understanding and calculating resistance is crucial because it tells you how well electricity can move through the solution.

Cell Constant

The cell constant is a crucial factor in determining the specific conductance of a solution. It defines the geometry of the electrodes in the conductivity cell used for measurement.
The cell constant is calculated using the formula:

• \( C = \frac{\text{Distance between electrodes}}{\text{Area of the electrodes}} \)

In practical terms, it connects the conductance observed in a cell to the conductance expected if the electrodes had a standard distance and area.

For most accurate results:

• The cell constant should be determined experimentally.
• It helps in standardizing measurements across different conductivity setups.

Using the cell constant ensures consistent and meaningful conductivity results, facilitating comparisons between different measurements.

Conductance

Conductance is the capacity of a solution to conduct electric current. It is the reciprocal of resistance and is measured in Siemens (\( \mathrm{S} \)).
In simple terms, conductance tells us how easily electricity can pass through a solution. If a solution's resistance is low, its conductance is high, suggesting it conducts electricity well.

In the given exercise:

• Conductance \( G \) is calculated as \( G = \frac{1}{R} \).
• This relationship highlights the interconnectedness between resistance and conductance.

Understanding conductance helps in contextualizing other electrochemical properties, aiding in calculation and analysis of how different solutions will behave under the influence of electric fields.