Question

A conductance cell when filled with 0.05 M solution of KCl records a resistance of 410.5 ohm at ${25}^{o}C$. When filled with $CaC{l}_2$ solution (11 g in 500 mL) it records 990 ohm. If the conductivity of 0.05 M KCl is $0.00189Sc{m}^{-1}$, the value of cell constant is:

• A. $0.667c{m}^{-1}$
• B. $0.677c{m}^{-1}$
• C. $0.776c{m}^{-1}$
• D. None of these

Answer 1

The correct option is C $0.776c{m}^{-1}$

The relationship between the conductivity $\kappa$, resistance $R$ and cell constant $b$ is :

$b=\kappa \times R$.

Substitute values in the above expression.

$b=0.00189S/cm\times 410.5ohm=0.776/cm$

Hence, the cell constant is $0.776/cm$.

Therefore, option C is correct.