Question

The specific conductance of a $0.1NKCl$ solution at ${23}^{\circ }C$ is $0.0112oh{m}^{-1}c{m}^{-1}$. The resistance of the cell containing the solution at the same temperature was found to be $55ohm$. The cell constant will be:

• A. $0.142c{m}^{-1}$
• B. $0.918c{m}^{-1}$
• C. $1.12c{m}^{-1}$
• D. $0.616c{m}^{-1}$

Answer 1

Answer: (D)

$0.616c{m}^{-1}$

$\kappa =G{G}^{\ast }$

where $\kappa =specificconductance=0.0112{ohm}^{-1}{cm}^{-1}$

G=conductance$=\frac{1}{55}=0.01818{ohm}^{-1}$

$G^{\ast }$=cell constant

$0.0112=0.01818\times G^{\ast }{G}^{\ast }=0.616{cm}^{-1}$