Question

The measured resistance of a conductance cell was 100 $\Omega$. ($M_{KCl}=$ $74.5gmo{l}^{-1}$ and cell constant $=1.25c{m}^{-1}$). Calculate the molar conductance of the solution.

(If the answer is $X$, then give an answer as $X-10.5$)

Answer 1

The expression for specific conductance $\kappa$ is

$\kappa =\frac{\frac{l}{a}}{R}$.

Here R is the resistance and $\frac{l}{a}$ is the cell constant.

Substitute values in the above expression.

$\kappa =\frac{1.25c{m}^{-1}}{100ohms}=0.0125oh{m}^{-1}c{m}^{-1}$.

Now to calculate the molar conductance, the concentration of the solution is needed.

Let the concentration of the solution be 1 M.

Hence, the molar conductivity of the solution will be

$\lambda =\frac{1000\kappa }{C}=\frac{1000c{m}^3{L}^{-1}\times 0.0125oh{m}^{-1}c{m}^{-1}}{1mol{L}^{-1}}=12.5Sc{m}^{2}mo{l}^{-1}$

Now,

$12.5-10.5=2$