Animated Solution for Chemistry - Electrochemistry: The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are $280$, $860$ and $426{\text{ S cm}}^2{\text{mol}}^{-1}$ respectively. The molar conductivity at infinite dilution of barium sulphate is ... ${\text{S cm}}^2{\text{mol}}^{-1}$ (Round off to the nearest Integer).

Imagine you are trying to measure the conductivity of a sparingly soluble salt like Barium Sulphate (${\text{BaSO}}_4$). Because it barely dissolves in water, you can't just plot a graph of its conductivity against concentration and extrapolate it to zero to find its limiting molar conductivity (${\Lambda }_m^{\infty }$). It's a classic roadblock in electrochemistry.

But here is where Kohlrausch's Law of Independent Migration of Ions comes to the rescue! This brilliant law states that at infinite dilution, where ions are so far apart they don't interact, the total molar conductivity of an electrolyte is simply the sum of the individual conductivities of its constituent ions. This means we can use strong electrolytes—which are easy to measure—as building blocks to construct the value for our stubborn sparingly soluble salt.

Let's look at the tools we have in our arsenal. We are given the limiting molar conductivities of three strong electrolytes:

According to Kohlrausch's law, we can break these down into their ionic components:

Our ultimate goal is to find the value for Barium Sulphate, which requires exactly one Barium ion and one Sulphate ion:

If we simply add the equations for ${\text{BaCl}}_2$ and ${\text{H}}_2{\text{SO}}_4$, we get our desired ${\text{Ba}}^{2+}$ and ${\text{SO}}_4^{2-}$ ions. However, we also accidentally invite two unwanted guests: $2{\text{H}}^{+}$ and $2{\text{Cl}}^{-}$.

How do we kick them out? By subtracting exactly two molecules of $\text{HCl}$! This gives us our master algebraic equation:

Now, it's just a matter of plugging in the numbers carefully. Don't rush the arithmetic!

First, multiply the $\text{HCl}$ value:

Next, add the first two terms:

Finally, subtract the unwanted ions:

And there we have it! The limiting molar conductivity of Barium Sulphate is $288{\text{ S cm}}^2{\text{mol}}^{-1}$. This elegant algebraic manipulation is a favorite concept in JEE, proving that sometimes, the best way to solve a complex physical chemistry problem is to treat it like a simple math puzzle.