Solveeit Logo
Login

Question

Question: The specific conductance of a \(0.1N\) \(KCl\) solution is \(0.012oh{m^{ - 1}}c{m^{ - 1}}\). The res...

The specific conductance of a 0.1N0.1N KClKCl solution is 0.012ohm−1cm−10.012oh{m^{ - 1}}c{m^{ - 1}}. The resistance of the cell containing the same solution is 55ohms55ohms. Then the cell constant is:
A. 0.918cm−10.918c{m^{ - 1}}
B. 0.66cm−10.66c{m^{ - 1}}
C. 1.142cm−11.142c{m^{ - 1}}
D. 1.12cm−11.12c{m^{ - 1}}

Explanation

Solution

We can solve this problem with the concept of specific conductivity. In simple words, conductance is the property of materials or solutions. Specific conductance is the estimation of ability of the solution or material to conduct electricity. It is known as “kappa”.

Complete step by step answer:
Conductance is the ability of material to flow the ions through itself and forms electricity. If the conductance of electricity occurs through the solution by the ions in the solution it is known as ionic conductance. Ionic conductance depends upon the nature of solution, numbers of ions in solution and temperature.
At any given time, specific conductance is the reciprocal of the specific resistance. Specific resistance is denoted as “rho” (ρ)(\rho ). Hence, the relationship between both the specific conductance and specific resistance can be shown as follows: κ=1ρ\kappa = \dfrac{1}{\rho } . Formula for specific conductance is given below:
κ=lAR\kappa = \dfrac{l}{{AR}} ,where ll is the distance between the electrode, AA is cross sectional area of the tank in which the solution kept, RR is the resistance of the solution.
The ratio of the ll and AA is called the cell constant which is denoted as G∗{G^*} and 1R\dfrac{1}{R} is conductance which is represented as GG . Hence, specific conductance can also be written as κ=G∗×G\kappa = {G^*} \times G.

In the given question, the specific conductance (κ)\left( \kappa \right) of a 0.1N0.1N KClKCl solution is 0.012ohm−1cm−10.012oh{m^{ - 1}}c{m^{ - 1}}. The resistance of the cell containing the same solution is 55ohms55ohms therefore conductance of the cell is 1R\dfrac{1}{R} =155=0.01818ohm−1 = \dfrac{1}{{55}} = 0.01818oh{m^{ - 1}} .
⇒0.012=0.01818×G∗\Rightarrow 0.012 = 0.01818 \times {G^*}
G∗=0.66cm−1{G^*} = 0.66c{m^{ - 1}}
Therefore, the cell constant is: G∗=0.66cm−1{G^*} = 0.66c{m^{ - 1}}. So, the correct answer is “Option B”.

Note: In the more scientifically word, specific conductance is the conductance of one unit volume of solution presented between two electrodes at a distance of unit length with the unit area of cross section.