Question

Equivalent conductance of an electrolyte containing $NaF$ at infinite dilution is $90.1Oh{m}^{-1}c{m}^2$. If $NaF$ is replaced by $KF$ what is the value of equivalent conductance?

• A. $90.1Oh{m}^{-1}c{m}^2$
• B. $11.2Oh{m}^{-1}c{m}^2$
• C. $0$
• D. $222.4Oh{m}^{-1}c{m}^2$

Answer 1

The correct option is A $90.1Oh{m}^{-1}c{m}^2$

At infinite dilution the equivalent conductance of strong electrolytes furnishing same number of ions is same.

For this case, Both NaF and KF are strong electrolyte and also furnish 2 ions in the solution.

Hence, Both will have same equivalent conductance.

So, option A is correct.