Q.The electrical resistance of a column of $0.05\ mol\ L^{-1}$ NaOH solution of diameter $1\ cm$ and length $50\ cm$ is $5.55 \times 10^{3}$ ohm. Calculate its resistivity, conductivity and molar conductivity.

Concept: Molar Conductivity — the conductivity of all ions produced by one mole of electrolyte in a given volume.

Step 1: Find the cross-sectional area.

Diameter $= 1\ \text{cm}$, so radius $r = 0.5\ \text{cm} = 0.005\ \text{m}$.

Area $A = \pi r^2 = \pi (0.005)^2 = 7.854 \times 10^{-5}\ \text{m}^2$.

Step 2: Calculate resistivity ($\rho$) using $R = \rho \frac{l}{A}$.

$\rho = R \cdot \frac{A}{l} = (5.55 \times 10^3) \cdot \frac{7.854 \times 10^{-5}}{0.50}$

$\rho = 5.55 \times 10^3 \times 1.5708 \times 10^{-4} = 0.872\ \Omega\ \text{m}$.

Step 3: Conductivity ($\kappa$) is the reciprocal of resistivity.

$\kappa = \frac{1}{\rho} = \frac{1}{0.872} = 1.147\ \text{S m}^{-1}$.

Step 4: Molar conductivity ($\Lambda_m$) = $\frac{\kappa}{c}$, with $c$ in $\text{mol/m}^3$. …

The key to solving this problem is understanding that resistance ($R$) is a property of the entire sample, while resistivity ($\rho$) and conductivity ($\kappa$) are intrinsic properties of the material. The geometry of the column — its length and cross-sectional area — acts as a bridge between them.

Think of it this way: a longer column offers more resistance (electrons have to travel further), and a wider column offers less resistance (more lanes for electrons to flow). The cell constant $G^*$ captures this geometry: $G^* = \frac{l}{A}$. Once you know $G^*$, you can extract the material's intrinsic conductivity from the measured resistance using $\kappa = \frac{G^*}{R}$.

Molar conductivity then extends this idea to account for concentration — it tells you how well a given amount of dissolved electrolyte conducts, making it possible to compare different solutions fairly.

Let’s work through it step by step.

1. Find the cross-sectional area of the column.

   The column is cylindrical with diameter $d = 1\ \text{cm}$, so radius $r = 0.5\ \text{cm}$.

   Area $A = \pi r^2 = \pi (0.5)^2 = 0.25\pi\ \text{cm}^2$.

   Numerically: $A = 0.25 \times 3.1416 \approx 0.7854\ \text{cm}^2$.

2. Calculate the cell constant $G^*$.

   The cell constant is defined as $G^* = \frac{l}{A}$, where $l = 50\ \text{cm}$ is the length between electrodes.

3. Compute resistivity $\rho$. Resistivity is related to resistance by $\rho = R \cdot \frac{A}{l} = \frac{R}{G^*}$. Given $R = 5.55 \times 10^3\ \Omega$,

4. Find conductivity $\kappa$. Conductivity is simply the reciprocal of resistivity: $\kappa = \frac{1}{\rho}$.

Alternatively, directly from the cell constant: $\kappa = \frac{G^*}{R} = \frac{63.66}{5.55 \times 10^3} = 0.011469\ \text{S cm}^{-1}$. Both routes give the same result.

5. Calculate molar conductivity $\Lambda_m$. …

Method: Resistance → Resistivity → Conductivity → Molar Conductivity

This is a step-by-step conversion method using the fundamental relations between resistance, cell geometry, and concentration.

Step 1: Find the cross-sectional area ($A$)

Given:

• Diameter $d = 1\ \text{cm} = 0.01\ \text{m}$
• Radius $r = \frac{d}{2} = 0.005\ \text{m}$

Step 2: Calculate resistivity ($\rho$)

Using the resistance formula:

Rearrange:

Given:

• $R = 5.55 \times 10^{3}\ \Omega$
• $l = 50\ \text{cm} = 0.5\ \text{m}$

Step 3: Calculate conductivity ($\kappa$)

Conductivity is the reciprocal of resistivity:

Step 4: Calculate molar conductivity ($\Lambda_m$)

Molar conductivity relates conductivity to concentration:

Given concentration:

• $c = 0.05\ \text{mol L}^{-1} = 0.05 \times 10^{3}\ \text{mol m}^{-3} = 50\ \text{mol m}^{-3}$

$\Lambda_m = \frac{1.147}{50}$ …

Here’s a breakdown of the common mistakes students make on this Molar Conductivity problem, and how to avoid each one.

1. Confusing Area with Radius (The $A = \pi r^2$ Trap)

The Mistake:

Students see “diameter = 1 cm” and plug $1$ directly into $A = \pi r^2$ as if it were the radius.

This gives $A = \pi (1)^2 = \pi$, which is wrong.

Why it happens:

The formula for area of a circle uses radius, not diameter. The problem gives diameter, so you must halve it first.

How to avoid:

• Always write: $r = \frac{\text{diameter}}{2}$
• Here: $r = \frac{1}{2} = 0.5\ \text{cm}$
• Then: $A = \pi r^2 = \pi (0.5)^2 = 0.25\pi\ \text{cm}^2$

Key rule: Diameter → radius → area. Never skip the halving step.

2. Forgetting Unit Conversion (cm → m)

The Mistake:

Using length in cm and area in cm² directly in the resistivity formula, then getting an answer off by a factor of $10^{-2}$ or $10^{-4}$.

Why it happens:

Resistivity ($\rho$) and conductivity ($\kappa$) are defined in SI units:

• Length in meters
• Area in meters²

How to avoid:

Convert before plugging into formulas:

• $l = 50\ \text{cm} = 0.50\ \text{m}$
• $r = 0.5\ \text{cm} = 0.005\ \text{m}$
• $A = \pi (0.005)^2 = \pi \times 25 \times 10^{-6} = 7.854 \times 10^{-5}\ \text{m}^2$

Tip: Write all given values in SI units at the very start of your solution.

3. Mixing Up Resistivity and Conductivity

The Mistake:

Thinking $\rho$ and $\kappa$ are the same, or using the wrong formula.

Why it happens:

They are reciprocals, but students often swap them or forget which is which.

How to avoid:

Memorise the relationship:

• Resistivity: $\rho = R \cdot \frac{A}{l}$
• Conductivity: $\kappa = \frac{1}{\rho} = \frac{l}{R \cdot A}$

Mnemonic: “Resistivity resists — it’s $R$ times $A/l$.”

4. Using Wrong Units for Molar Conductivity

The Mistake:

Writing molar conductivity in $\text{S cm}^2 \text{mol}^{-1}$ but forgetting to convert conductivity from $\text{S m}^{-1}$ to $\text{S cm}^{-1}$.

Why it happens:

Concentration is given in $\text{mol L}^{-1}$, but the standard unit for molar conductivity is $\text{S cm}^2 \text{mol}^{-1}$.

If $\kappa$ is in $\text{S m}^{-1}$, you must convert:

$1\ \text{S m}^{-1} = 0.01\ \text{S cm}^{-1}$

How to avoid:

Use the formula:

Or convert $\kappa$ to $\text{S cm}^{-1}$ first by dividing by 100.

Check: If your $\Lambda_m$ comes out as a very small number (like $10^{-6}$), you likely forgot the conversion.

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