Q.Conductivity $\kappa$, is equal to ____________. (Two or more than two options may be correct.)

The key idea is Molar Conductivity — but the question is about conductivity $\kappa$, not molar conductivity. Conductivity is the reciprocal of resistivity, and resistivity is $R \frac{A}{l}$.

Step 1: From Ohm’s law, resistance $R = \rho \frac{l}{A}$, so resistivity $\rho = R \frac{A}{l}$.

Step 2: Conductivity $\kappa = \frac{1}{\rho} = \frac{1}{R} \frac{l}{A}$. This matches option (i).

Step 3: The cell constant $G^* = \frac{l}{A}$, so $\kappa = \frac{G^*}{R}$. This matches option (ii). …

The key here is to remember what conductivity actually means physically. Conductivity ($\kappa$) is a measure of how easily a solution conducts electricity — it’s the conductance of a 1 cm cube of the solution. That’s the intuitive picture: imagine a cube of the electrolyte, 1 cm on each side; the conductance you measure across opposite faces is the conductivity.

Now, how do we get there from measurable quantities? In the lab, you measure resistance $R$ of a solution placed in a cell with two electrodes of area $A$ separated by distance $l$. The relationship between resistance and the geometry is:

where $\rho$ is resistivity. Since conductivity is the reciprocal of resistivity:

That’s option (i) directly.

But there’s a practical twist. In a real conductivity cell, you rarely know $l$ and $A$ separately — instead, you calibrate the cell and determine its cell constant $G^* = \frac{l}{A}$. So the same formula becomes:

That’s option (ii).

Now let’s check the others:

1. Option (i): $\frac{1}{R}\frac{l}{A}$ — This is exactly the definition from $R = \rho \frac{l}{A}$ and $\kappa = 1/\rho$. So this is correct.

2. Option (ii): $\frac{G^*}{R}$ — Since $G^* = l/A$, this is just a repackaging of (i). Also correct. …

Method: Direct Formula Application for Conductivity ($\kappa$)

This method uses the definition of conductivity in terms of resistance and cell geometry.

Step 1: Recall the basic relation

Conductivity $\kappa$ is the reciprocal of resistivity ($\rho$):

Step 2: Express resistivity in terms of resistance and cell dimensions

Resistivity is given by:

where:

• $R$ = resistance
• $A$ = cross-sectional area of electrodes
• $l$ = distance between electrodes

Step 3: Substitute to get conductivity

This matches option (i).

Step 4: Introduce cell constant

The term $\frac{l}{A}$ is called the cell constant, $G^*$:

Thus:

This matches option (ii). …

The Core Concept First

Conductivity ($\kappa$) is the ability of a solution to conduct electricity. It is defined as:

where $\rho$ is resistivity.

From the resistance formula:

we get:

Also, the cell constant $G^* = \frac{l}{A}$, so:

Correct Answer

✓ Options (i) and (ii) are correct.

• (i) $\frac{1}{R} \frac{l}{A}$ — direct formula.
• (ii) $\frac{G^*}{R}$ — using cell constant.

✗ (iii) $\Lambda_m$ is molar conductivity, not conductivity.

✗ (iv) $\frac{l}{A}$ is the cell constant, not conductivity.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing $\kappa$ with $\Lambda_m$

• Why it happens: Both are measures of conductance, and symbols look similar.
• How to avoid: Remember:
  • $\kappa$ = conductivity (S cm⁻¹ or S m⁻¹)
  • $\Lambda_m$ = molar conductivity (S cm² mol⁻¹ or S m² mol⁻¹)
  • Relation: $\Lambda_m = \frac{\kappa}{c}$ (where $c$ is concentration)

Mistake 2: Thinking $\frac{l}{A}$ alone is conductivity

• Why it happens: Students see $l/A$ in the formula and forget the $1/R$ factor.
• How to avoid: Always write the full definition:

The $l/A$ part is just the cell constant, not conductivity.

Mistake 3: Forgetting the reciprocal relationship

• Why it happens: Mixing up $R = \rho \frac{l}{A}$ with $\kappa = \frac{1}{\rho}$.
• How to avoid: Draw a quick memory chain: …