Animated Solution for Chemistry - Electrochemistry: A $5.0{\text{ m mol dm}}^{-3}$ aqueous solution of KCl has a conductance of $0.55\text{ mS}$ when measured in a cell constant $1.3{\text{ cm}}^{-1}$. The molar conductivity of this solution is .......... ${\text{mS m}}^2{\text{ mol}}^{-1}$. (Round off to the nearest integer)

Imagine you are working in a chemistry lab, and you have a conductivity cell filled with a dilute aqueous solution of potassium chloride ($\text{KCl}$). The problem provides us with three key pieces of information: the concentration of the solution ($C=5.0{\text{ m mol dm}}^{-3}$), the measured conductance ($G=0.55\text{ mS}$), and the cell constant ($G^{\ast }=1.3{\text{ cm}}^{-1}$). Our ultimate goal is to determine the molar conductivity (${\Lambda }_m$) of this solution in the specific units of ${\text{mS m}}^2{\text{ mol}}^{-1}$.

First, let's recall the fundamental relationship between conductivity ($\kappa$), conductance ($G$), and the cell constant ($G^{\ast }$). Conductivity is an intensive property of the solution, and it is simply the product of the measured conductance and the cell constant:

Before we blindly substitute the numbers, we must watch out for a classic trap: units. The cell constant is given in inverse centimeters (${\text{cm}}^{-1}$), but our final answer requires inverse meters (${\text{m}}^{-1}$). Let's convert it right away to avoid any mismatch later. Since $1\text{ cm}=10^{-2}\text{ m}$, it follows that $1{\text{ cm}}^{-1}=100{\text{ m}}^{-1}$. Therefore:

Now, we can safely substitute the values into our master equation. Multiplying the conductance by the converted cell constant gives us the conductivity:

Next, we need to find the molar conductivity (${\Lambda }_m$), which is defined as the conductivity divided by the concentration of the solution:

Here is where many students make a silly mistake. The concentration is given as $5.0{\text{ m mol dm}}^{-3}$. We need to convert this into standard SI units of ${\text{mol m}}^{-3}$ to match our conductivity units. Let's break it down: "m mol" means $10^{-3}\text{ mol}$, and "${\text{dm}}^{-3}$" means per liter, which is $10^{-3}{\text{ m}}^3$.

When we put it together, the $10^{-3}$ factors in the numerator and denominator cancel out perfectly! So, $5.0{\text{ m mol dm}}^{-3}$ is exactly equal to $5.0{\text{ mol m}}^{-3}$. This is a beautiful simplification!

With both $\kappa$ and $C$ in the correct units, we are ready for the final calculation. Let's substitute them into the molar conductivity formula:

The question explicitly asks us to round off the final answer to the nearest integer. Since $14.3$ is closer to $14$ than $15$, we round it down.

Final Answer: The molar conductivity of the solution is 14.