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 is \(5 \times 10^2\) \(\text{S m}^2 \text{mol}^{-1}\) (option b).

Step 1: Understanding the Problem

We need to calculate the molar conductivity of a 0.5 M solution. We have the resistance of the solution and the specific conductance.

Step 2: Calculate the Cell Constant

The cell constant, denoted as \( K \), can be calculated using the formula \( K = \text{specific conductance} \times \text{resistance} \) for the 0.2 M solution.\[ K = 1.4 \, \text{S m}^{-1} \times 50 \, \Omega = 70 \, \text{m}^{-1} \]

Step 3: Determine Conductance of 0.5 M Solution

The conductance \((C)\) for the 0.5 M solution is given by \( C = \frac{1}{\text{resistance}} \).\[ C = \frac{1}{280 \, \Omega} = \frac{1}{280} \, \text{S} \approx 0.00357 \, \text{S} \]

Step 4: Calculate Specific Conductance of 0.5 M Solution

Use the cell constant from step 2 and the conductance from step 3 to find the specific conductance \((k)\). Use the formula \( k = \frac{K}{R} \).Substituting the known values:\[ k = 70 \, \text{m}^{-1} \times 0.00357 \, \text{S} = 0.25 \, \text{S m}^{-1} \]

Step 5: Calculate Molar Conductivity

Molar conductivity \((\Lambda_m)\) is calculated as \( \Lambda_m = \frac{k}{C} \) where \( C \) is the concentration.\[ \Lambda_m = \frac{0.25 \, \text{S m}^{-1}}{0.5 \, \text{mol m}^{-3}} = 0.5 \, \text{S m}^2 \text{mol}^{-1} \]

Step 6: Convert to Desired Units

0.5 \(\text{S m}^2 \text{mol}^{-1}\) is equivalent to \(5 \times 10^2\text{ S cm}^2 \text{mol}^{-1}\), matching the units needed for the answer.

Key Concepts

Specific Conductance

Specific conductance, also known as conductivity, is a measure of a solution's ability to conduct electricity. It is denoted by the symbol \( k \) and is expressed in Siemens per meter \( \text{S m}^{-1} \).
It helps us understand how easily electricity can flow through a given electrolyte solution.When we deal with an electrolyte solution, ions are present. These ions are the charge carriers that facilitate electrical conduction through the solution.
As the concentration of ions increases, typically, the specific conductance increases. This is because more ions can carry more electric current.

• The formula to find specific conductance is \( k = \frac{K}{R} \), where \( K \) is the cell constant, and \( R \) is the resistance of the solution.

This relationship highlights how specific conductance can change based on both the cell's geometry (as represented by the cell constant) and the solution's resistance.

Cell Constant

The cell constant \( (K) \) is a critical factor when measuring the conductive properties of ions in a solution. It is inversely proportional to both the length and directly proportional to the area of the electrodes.
This constant makes it possible to translate between the physical geometry of the measurement cell and the solution's conductivity.In easier terms, the cell constant helps adjust any measurements you take of the solution.

• Formula to calculate cell constant is \( K = \text{specific conductance} \times \text{resistance} \).

In our exercise, we computed the cell constant for a 0.2 M solution using the formula \( K = 1.4 \, \text{S m}^{-1} \times 50 \, \Omega = 70 \, \text{m}^{-1} \).
The calculated value helps us understand the intrinsic factor that adjusts the raw measurements we make.

Electrolyte Solution

An electrolyte solution is typically a liquid containing ions, which are mobile, charged particles. These ions make the solution conductive.
Commonly, salts and their solutions in water act as electrolytes because they can dissolve into positive and negative ions. The concentration of the electrolyte is directly related to its conductivity.

• For example, a 0.5 M solution is more concentrated than a 0.2 M solution.
• In our scenario, different molarities have shown different resistances and conductivities due to varying ion concentrations.

Knowing about electrolyte solutions helps to predict how they will behave under electrical currents, crucial for applications like batteries or in designing better conductors.

Resistance Calculation

Resistance, denoted by \( R \), is the property of a material that opposes the flow of electric current.
It is measured in ohms \((\Omega)\). In an electrolyte solution, resistance depends on factors like ion concentration and the nature of the electrolyte.To find resistance in relation to the conductivity of an electrolyte, you can use the mathematical expression:

• \( R = \frac{1}{C} \)

Here, \( C \) is the conductance, which is the reciprocal of resistance.
In our example, for a 0.5 M solution, we've determined its resistance as \( 280 \, \Omega \) from which we calculated conductance as about 0.00357 S.
These calculations guide us to understand how resistant or conductive an electrolyte solution is, aiding in the proper adjustments needed in practical applications involving electric currents.