Question

The conductivity of N/50 solution of KCl in a cell at ${25}^{o}C$ is $0.002765Sc{m}^{-1}$. If the resistance of a cell containing this solution is 400 ohm, cell constant will be:

• A. $1.106c{m}^{-1}$
• B. $3.101c{m}^{-1}$
• C. $6.106c{m}^{-1}$
• D. None of these

Answer 1

The correct option is A $1.106c{m}^{-1}$

Given for (N/50) solution
$k=0.002765Sc{m}^{-1},R=400ohm$

Now,

$k=\frac{1}{R}\times \frac{l}{a}$

$0.002765=\frac{1}{400}\times$ cell constant

$\therefore$ Cell constant $=1.106c{m}^{-1}$