Q.Resistance of a conductivity cell filled with $0.1\ mol\ L^{-1}$ KCl solution is $100\ \Omega$. If the resistance of the same cell when filled with $0.02\ mol\ L^{-1}$ KCl solution is $520\ \Omega$, calculate the conductivity and molar conductivity of $0.02\ mol\ L^{-1}$ KCl solution. The conductivity of $0.1\ mol\ L^{-1}$ KCl solution is $1.29\ S\ m^{-1}$.

Concept: Conductance and Conductivity — the cell constant $G^*$ links measured resistance $R$ to conductivity $\kappa$ via $\kappa = G^*/R$.

Step 1: Find the cell constant

For the $0.1\ \text{mol L}^{-1}$ solution:

$\kappa_1 = 1.29\ \text{S m}^{-1}$, $R_1 = 100\ \Omega$

Step 2: Conductivity of $0.02\ \text{mol L}^{-1}$ solution

$R_2 = 520\ \Omega$

Step 3: Molar conductivity

Concentration $c = 0.02\ \text{mol L}^{-1} = 20\ \text{mol m}^{-3}$

Why this works: Conductance and conductivity

The key idea is that a conductivity cell has a fixed geometry — the distance between electrodes and their area don't change. This geometry is captured by the cell constant, $G^*$, with units of $\mathrm{m^{-1}}$:

Conductance $G$ (in siemens, S) is the reciprocal of resistance: $G = 1/R$. Conductivity $\kappa$ (in $\mathrm{S\ m^{-1}}$) is related to conductance by:

So if we know $\kappa$ and $R$ for one solution, we can find $G^*$. Then for any other solution in the same cell, knowing $R$ gives $\kappa$.

Molar conductivity $\Lambda_m$ (in $\mathrm{S\ m^2\ mol^{-1}}$) is then:

where $c$ is concentration in $\mathrm{mol\ m^{-3}}$. This is the conductivity per mole of electrolyte — it tells us how well each mole carries current.

Step-by-step solution

1. Find the cell constant using the 0.1 M KCl data.

We are given:

• For $0.1\ \mathrm{mol\ L^{-1}}$ KCl: $R_1 = 100\ \Omega$, $\kappa_1 = 1.29\ \mathrm{S\ m^{-1}}$

From $\kappa = G^*/R$, we get:

2. Calculate the conductivity of the 0.02 M KCl solution.

For the same cell, $G^*$ is fixed. With $R_2 = 520\ \Omega$:

3. Convert concentration to SI units ($\mathrm{mol\ m^{-3}}$).

The given concentration is $0.02\ \mathrm{mol\ L^{-1}}$. Since $1\ \mathrm{L} = 10^{-3}\ \mathrm{m^3}$:

4. Compute molar conductivity.

In scientific notation:

Method: Cell Constant Method

This method uses the cell constant ($G^*$) of the conductivity cell, which remains fixed for a given cell. The cell constant relates resistance ($R$) to conductivity ($\kappa$) via:

Step 1: Find the cell constant using the known solution

For the $0.1\ \text{mol L}^{-1}$ KCl solution:

• Given: $R_1 = 100\ \Omega$, $\kappa_1 = 1.29\ \text{S m}^{-1}$

Using the formula:

Cell constant $G^* = 129\ \text{m}^{-1}$

Step 2: Calculate conductivity of the $0.02\ \text{mol L}^{-1}$ KCl solution

For the unknown solution:

• Given: $R_2 = 520\ \Omega$
• Cell constant is the same: $G^* = 129\ \text{m}^{-1}$

Conductivity $\kappa_2 = \mathbf{0.248\ S\ m^{-1}}$

Step 3: Calculate molar conductivity of the $0.02\ \text{mol L}^{-1}$ KCl solution

Molar conductivity ($\Lambda_m$) is given by:

where:

• $\kappa$ is in $\text{S m}^{-1}$
• $c$ is concentration in $\text{mol m}^{-3}$

Convert concentration:

Now:

Molar conductivity $\Lambda_m = \mathbf{1.24 \times 10^{-2}\ S\ m^2\ mol^{-1}}$

Final Answer

Here are the most common mistakes students make with this problem, along with the conceptual fixes to avoid them.

1. Forgetting the Cell Constant is the Bridge

The Mistake: Students try to directly use the formula $\kappa = \frac{1}{R} \times \frac{l}{A}$ without first calculating the cell constant ($G^* = l/A$) from the known data.

Why it happens: They see two resistances and two concentrations and panic, trying to plug numbers into the wrong formula.

How to Avoid:

• Concept: The cell constant ($l/A$) is a property of the physical cell (the distance between electrodes and their area). It does not change when you change the solution.
• Action: Always calculate $G^*$ first using the data for the known solution ($0.1\ \text{mol L}^{-1}$ KCl).

2. Unit Confusion (cm vs. m)

The Mistake: Using $\kappa$ in $\text{S cm}^{-1}$ when the problem gives $\kappa$ in $\text{S m}^{-1}$, or forgetting to convert concentration from $\text{mol L}^{-1}$ to $\text{mol m}^{-3}$ for molar conductivity.

Why it happens: Electrochemistry problems often mix units. Conductivity is often given in $\text{S cm}^{-1}$ in textbooks, but here it's in $\text{S m}^{-1}$.

How to Avoid:

• Check units at the start. The given $\kappa = 1.29\ \text{S m}^{-1}$.
• Cell constant will be in $\text{m}^{-1}$ (since $R$ is in $\Omega$).
• Conductivity of the unknown ($\kappa_{0.02}$) will come out in $\text{S m}^{-1}$.
• For molar conductivity ($\Lambda_m$): Convert concentration from $\text{mol L}^{-1}$ to $\text{mol m}^{-3}$.

3. Using the Wrong Resistance for the Cell Constant

The Mistake: Using the resistance of the $0.02\ \text{mol L}^{-1}$ solution ($520\ \Omega$) to calculate the cell constant.

Why it happens: Students think "cell constant" is calculated from the solution they are trying to find, not from the standard/reference solution.

How to Avoid:

• Rule: The cell constant is always calculated from the solution whose conductivity is known.
• Here, the known solution is $0.1\ \text{mol L}^{-1}$ KCl with $\kappa = 1.29\ \text{S m}^{-1}$ and $R = 100\ \Omega$.
• The $0.02\ \text{mol L}^{-1}$ solution is the unknown — you use its resistance after you have $G^*$.

4. Confusing Conductivity ($\kappa$) with Conductance ($G$)

The Mistake: Thinking that $1/R$ (conductance) is the same as conductivity ($\kappa$).

Why it happens: The words sound similar, and both involve resistance.

How to Avoid:

• Remember the relationship:

where $G = 1/R$.

• Conductivity ($\kappa$) is conductance per unit length and area — it's an intensive property of the solution.
• Conductance ($G$) depends on the geometry of the cell.

5. Forgetting the Final Step: Molar Conductivity

The Mistake: Stopping after finding conductivity ($\kappa$) and not calculating molar conductivity ($\Lambda_m$).

Why it happens: The question explicitly asks for both conductivity and molar conductivity, but students rush.

How to Avoid:

• Read the question twice. Underline "conductivity and molar conductivity".
• Formula:

where $c$ is in $\text{mol m}^{-3}$.

• Plug in carefully:

Quick Summary Checklist

Final Correct Values (for your reference):

• Cell constant: $129\ \text{m}^{-1}$
• Conductivity of $0.02\ \text{M}$ KCl: $\mathbf{0.248\ \text{S m}^{-1}}$
• Molar conductivity: $\mathbf{1.24 \times 10^{-2}\ \text{S m}^2 \text{mol}^{-1}}$