Question

Specific conductance of $\dfrac{N}{50}$ KCl solution at $25{}^\circ C$is 0.002765 mho$c{{m}^{-1}}$. If the resistances of cell along with the solution is 400 ohm, then cell constant will be:
(A) 0.553$c{{m}^{-1}}$
(B) 1.106$c{{m}^{-1}}$
(C) 2.212$c{{m}^{-1}}$
(D) None of these

Answer 1

Hint:The specific conductance of a solution is equal to the ratio of the cell constant of the solution to the resistance of the cell. It is denoted by the symbol ‘k’. The formula for cell constant of a cell is given by:
Cell constant (a) = $k\times R$
Where R is the resistance of the cell

Complete step by step solution:
Let’s look at the solution of the question:
The values given in the question are:
‘k’ = 0.002765 mho$c{{m}^{-1}}$
R = 400 ohm
Now, we will calculate the cell constant of the given cell using the formula:
Cell constant (a) = $k\times R$
$a\,=\,0.002765\,\times \,400$
$a\,=\,1.106\,c{{m}^{-1}}$
Where, ‘a’ is the cell constant
Hence, the answer to the given question is option (B).

Additional Information:
Cell constant is the ratio of the distance between the electrodes of the cell to the area of electrodes.
The formula for cell constant is given by
$a\,=\,\dfrac{l}{A}$ where l is the distance between the electrodes and A is the area of electrodes.

Note: Conductance of a solution is the reciprocal of the resistance of the solution. The solution of KCl is used as a reference solution for the calibration of various conductance cells.