Question

The specific conductance of a $N / 10 \,KCl$ at $25^{\circ} C$ is $0.0112\, ohm ^{-1} cm ^{-1} .$ The resistance of cell containing solution at the same temperature was found to be $55\, ohm$. The cell constant will be

• A. 0.142 $cm^{-1} $
• B. 0.66 $cm^{-1} $
• C. 0.918 $cm^{-1} $
• D. 1.12 $cm^{-1} $

Answer 1

Answer: (B)

Specific conductivity, $k=0.012\, \Omega^{-1} \,cm ^{-1}$
$k=\frac{1}{\text { resistance }} \times \frac{1}{a}\left[G=\frac{1}{R}\right] $
$\frac{1}{a}=$ cell constant
$\frac{1}{a}=55 \times 0.012=0.66\, cm ^{-1}$