The resistance of a conductivity cell with cell constant 1.14 cm^-1, containing 0.001 M KCl at 298 K

Q: Why do we multiply by 1000 in the molar conductivity formula?

The factor of 1000 is a unit conversion factor. Molarity is given in moles per liter (or $\text{mol dm}^{-3}$). To convert this to $\text{mol cm}^{-3}$ so it matches the $\text{cm}^{-1}$ in specific conductivity, we divide the concentration by 1000, which mathematically brings the 1000 to the numerator.

Q: What is the difference between specific conductivity and molar conductivity?

Specific conductivity ($\kappa$) is the conductance of 1 $\text{cm}^3$ of the solution. Molar conductivity ($\Lambda_m$) is the conducting power of all the ions produced by dissolving 1 mole of an electrolyte in a given volume of solution.

Q: Can I convert the cell constant to meters first?

You can, but it will complicate things. The question specifically asks for the final answer in $\text{S cm}^2\text{mol}^{-1}$. Since the cell constant is already in $\text{cm}^{-1}$, it is much faster and less error-prone to stick to CGS units throughout the calculation.

Q: Why did we not calculate the exact decimal value of $\kappa$?

Calculating the exact decimal value of $\kappa = \frac{1.14}{1500}$ gives a very small number ($0.00076$). Working with such small decimals increases the risk of calculation errors. Keeping it as a fraction allows for easy cancellation in the final step.

Q: What are the units of cell constant?

The cell constant ($G^*$) is defined as the ratio of the distance between the electrodes ($l$) to their area of cross-section ($A$). Therefore, its units are $\text{length} / \text{length}^2$, which simplifies to $\text{length}^{-1}$ (e.g., $\text{cm}^{-1}$ or $\text{m}^{-1}$).

Question

The resistance of a conductivity cell with cell constant $1.14 	ext{ cm}^{-1}$, containing $0.001 	ext{ M KCl}$ at $298 	ext{ K}$ is $1500 \ \Omega$. The molar conductivity of $0.001 	ext{ M KCl}$ solution at $298 	ext{ K}$ in $	ext{S cm}^2	ext{mol}^{-1}$ is .......... . (Integer answer)

Sigma Percentile · Accepted Answer

Welcome to this classic electrochemistry problem! Imagine you are in a lab, and you have a conductivity cell right in front of you. We've filled it with a $0.001 	ext{ M}$ potassium chloride solution. We hook it up to a meter and find its resistance is exactly $1500 \ \Omega$. The manufacturer tells us the cell constant is $1.14 	ext{ cm}^{-1}$. Our mission? To find the molar conductivity of this solution.

### Finding the Specific Conductivity

To reach molar conductivity, we first need a stepping stone: the specific conductivity, which we call $\kappa$. Think of $\kappa$ as the true conducting power of the solution, independent of the cell's dimensions. How do we find it? Well, $\kappa$ is simply the cell constant, $G^*$, divided by the measured resistance, $R$. It's that straightforward!

Let's bring in our numbers. We substitute $1.14$ for the cell constant and $1500$ for the resistance.

$$\kappa = \frac{1.14}{1500} 	ext{ S cm}^{-1}$$

Now, here is a pro tip: don't rush to your calculator to divide this just yet! Silly mistakes happen when we deal with tiny decimals. Let's keep it as a neat fraction for now. It will make our lives much easier in the next steps.

### The Master Formula for Molar Conductivity

Now for the main event! The master formula for molar conductivity, $\Lambda_m$. This formula bridges the gap between the specific conductivity we just found and the concentration of our solution. Because we want our final answer in standard CGS units, $	ext{S cm}^2	ext{mol}^{-1}$, we multiply $\kappa$ by $1000$ and divide by the molarity, $C$.

$$\Lambda_m = \frac{\kappa 	imes 1000}{C}$$

Alright, let's carefully plug everything in. We substitute our fraction for $\kappa$, and $0.001$ for the concentration $C$.

$$\Lambda_m = \frac{\left(\frac{1.14}{1500}ight) 	imes 1000}{0.001}$$

Look at this expression. It might look a bit heavy, but notice how keeping $\kappa$ as a fraction is setting us up for some beautiful cancellations. This is why patience pays off in physical chemistry!

### The Final Calculation

Let's simplify this step by step. In the numerator, $1.14$ multiplied by $1000$ shifts the decimal three places, giving us $1140$. In the denominator, $1500$ multiplied by $0.001$ shifts the decimal back, leaving us with a very manageable $1.5$. See? No messy decimals!

$$\Lambda_m = \frac{1140}{1.5}$$

We are at the finish line! Dividing $1140$ by $1.5$ gives us exactly $760$.

**$\Lambda_m = 760 	ext{ S cm}^2	ext{mol}^{-1}$**

So, the molar conductivity of our potassium chloride solution is $760 	ext{ S cm}^2	ext{mol}^{-1}$. A perfect integer answer, just as the question demanded. Before we move on, a quick word of caution. This is a JEE favorite concept, and the trap is always in the units! The factor of $1000$ in our formula is specifically because our cell constant was in $	ext{cm}^{-1}$ and we wanted the answer in CGS units. If the units were in meters, the formula changes. Always keep your eyes wide open for units!

Animated Solution for Chemistry - Electrochemistry: The resistance of a conductivity cell with cell constant $1.14 cm^{- 1}$ , containing $0.001 M KCl$ at $298 K$ is $1500 Ω$ . The molar conductivity of $0.001 M KCl$ solution at $298 K$ in $S cm^{2} mol^{- 1}$ is .......... . (Integer answer)

Enter Numerical Value:

Visualized Solution

\text{The Conductivity Cell}

\text{Specific Conductivity } (\kappa)

\text{Substituting Values for } \kappa

\text{Molar Conductivity } (\Lambda_m)

\text{Substituting Values for } \Lambda_m

\text{Simplifying the Expression}

\text{Final Calculation}

\text{What if units change?}

The Sigma Insight: Electrolytic Conduction

Finding the Specific Conductivity

The Master Formula for Molar Conductivity

The Final Calculation

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