Question

Specific conductance of $0.1 \,N \,KCl$ solution at $23^{\circ} C$ is $0.012 \,ohm ^{-1} \,cm ^{-1}$. Resistance of cell containing the solution at same temperature was found to be $55 \,ohm$. The cell constant is:

• A. $0.0616\, cm ^{-1}$
• B. $0.616 \,cm ^{-1}$
• C. $6.16 \,cm ^{-1}$
• D. $616 \,cm ^{-1}$

Answer 1

Answer: (B)

$\kappa= GG ^{*}$
where $\kappa=$ specific
conductance $=0.0112\, ohm ^{-1} \,cm ^{-1}$
$G =$ conductance $=\frac{1}{55}=0.01818 \,ohm ^{-1}$
$G ^{*}=$ cell constant
$0.0112=0.01818 \times G ^{*}$
$G ^{*}=0.616\, cm ^{-1}$