Question

The cell constant of a given cell is $0.45c{m}^{-1}$. The resistance of 0.1 molar $K_4\left[Fe\right(CN{)}_6]$ solution placed in this cell is measured to be $36.0\Omega$ . What is the
equivalent conductivity of the solution (in $Sc{m}^{2}eq{v}^{-1}$)?

Answer 1

Conductance $=\frac{1}{resistance}=\frac{1}{36}S$

Conductivity = conductance $\times$ cell constant
$\Rightarrow Conductivity=\frac{1}{36}\times 0.45$
Equivalent conductivity $=\frac{k\times 1000}{N}=\frac{\frac{0.45}{36}\times 1000}{0.1\times 4}=31.25Sc{m}^{2}eq{v}^{-1}$