Question

The specific conductance of a $\frac{N}{10}KCl$ solution at ${18}^{0}C$ is $1.12\times {10}^{-2}mhoc{m}^{-1}$. The resistance of the soution contained in the cell is found to be $65ohm$. Calculate the cell constant:

• A. $0.912c{m}^{-1}$
• B. $0.512c{m}^{-1}$
• C. $0.728c{m}^{-1}$
• D. $0.632c{m}^{-1}$

Answer 1

The correct option is C $0.728c{m}^{-1}$

As given,
Specific conductance$\left(k\right)=1.12\times {10}^{-2}mhoc{m}^{-1}$
$\frac{N}{10}KCl$ solution
Resistance$\left(R\right)=65ohm$
We know that,
$k=\frac{1}{R}\times \frac{l}{a}$ $(\therefore \frac{l}{a}=\text{cell constant})$
$\Rightarrow \frac{l}{a}=k\times R$

$\Rightarrow \frac{l}{a}=1.12\times {10}^{-2}\times 65$

$\Rightarrow \frac{l}{a}=0.728c{m}^{-1}$