Question

Resistance and conductivity of a cell containing $ {\text{0}}{\text{.001}} $ M KCl solution at 298K are 1500 $ {\text{\Omega }} $ and $ {\text{1}}{\text{.46 \times 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{S}}{\text{.c}}{{\text{m}}^{{\text{ - 1}}}} $ respectively. What is the cell constant?
(A) $ {\text{0}}{\text{.219c}}{{\text{m}}^{{\text{ - 1}}}} $
(B) $ {\text{0}}{\text{.319c}}{{\text{m}}^{{\text{ - 1}}}} $
(C) $ {\text{0}}{\text{.419c}}{{\text{m}}^{{\text{ - 1}}}} $
(D) None of these

Answer 1

Hint: In the above question, the resistance and conductivity of a cell containing $ {\text{0}}{\text{.001}} $ M KCl solution at 298K is given. We have to find out the cell constant which is the ratio of conductivity and conductance, here conductance is the reciprocal of resistance.

Formula used:
Cell constant = $ \dfrac{{\text{k}}}{{\text{G}}} $
Where k is the conductivity
G is the conductance.

Complete step by step solution:
In the above equation, we have to find out the cell constant when conductivity and resistance is given. Cell constant is simply a number which denotes the ratio of the distance between the electrode plates to the surface area of the plate.
We know that cell constant = $ \dfrac{{\text{k}}}{{\text{G}}} $
G is the inverse of R. Hence, we can write cell constant as:
Cell constant = $ {\text{k \times R}} $
Substituting the values of k and R in the above equation, we get that:
Cell constant = $ {\text{k \times R = 1}}{\text{.46 \times 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ \times 1500 = 0}}{\text{.219c}}{{\text{m}}^{{\text{ - 1}}}} $
Since, cell constant is equal to $ {\text{0}}{\text{.219c}}{{\text{m}}^{{\text{ - 1}}}} $ .
Hence, the correct option is option A.

Note:
We know that resistance depends upon resistivity of the material, length and area of the material. Mathematically, it can be written as:
 $ {\text{R = \rho }}\dfrac{{\text{l}}}{{\text{A}}} $
Where R is the resistance
 $ {\text{\rho }} $ = resistivity
l= length of the conductor
A = area of the conductor
Conductance is the opposite of resistance.
 $ {\text{G = }}\dfrac{{\text{1}}}{{\text{R}}} $
Substituting the value of R, we get:
 $ {\text{G = }}\dfrac{{\text{A}}}{{{\text{\rho l}}}} $
Which implies that $ {\text{G = }}\dfrac{{\text{k}}}{{{\text{cell constant}}}} $
Where k is the conductivity which is inverse of resistivity and cell constant is $ \dfrac{{\text{l}}}{{\text{A}}} $ .