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Question: The resistance of the conductivity cell containing \(0.001M\) \(KCl\) solution at \(298K\) is \(1500...

The resistance of the conductivity cell containing 0.001M0.001M KClKCl solution at 298K298K is 1500Ω1500\Omega . Find the value of cell constant if conductivity of 0.001M0.001M KClKCl solution at 298K298K is 0.146×10−3S/cm0.146 \times {10^{ - 3}}S/cm

Explanation

Solution

From the definition of conductance, conductance is the reciprocal of the resistance. Then, cell constant can be calculated as the ratio of conductivity to the conductance.

Complete answer:
In this question, we have to find the cell constant, so we need to know something about cell constant, so, the cell constant can be defined as the function of electrode areas, the distance between the electrodes and electrical field pattern between electrodes. It can also be defined as the ratio of conductivity to conductance. It is denoted by xx Mathematically, it can be represented as –
⇒x=kG\Rightarrow x = \dfrac{k}{G}
where, kk is the conductivity, and
GG is the conductance
Now, in the question, let the resistance be RR.
According to the question, it is given that –
Resistance, R=1500ΩR = 1500\Omega , and
Conductivity, k=0.146×10−3S/cmk = 0.146 \times {10^{ - 3}}S/cm
Now, we have to find the conductance, therefore, from the definition of conductance, it can be defined as an ability of an electrolyte to conduct electric current. Therefore, the conductance will be –
⇒1R=11500\Rightarrow \dfrac{1}{R} = \dfrac{1}{{1500}}
Now, we know that, cell constant can be calculated as –
x=kGx = \dfrac{k}{G}
Putting the values of conductance and the conductivity in the above formula of cell constant, we get –
⇒x=0.146×10−311500 ⇒x=1500×0.146×10−3  \Rightarrow x = \dfrac{{0.146 \times {{10}^{ - 3}}}}{{\dfrac{1}{{1500}}}} \\\ \Rightarrow x = 1500 \times 0.146 \times {10^{ - 3}} \\\
By doing further calculations, we get –
⇒x=0.219cm−1\Rightarrow x = 0.219c{m^{ - 1}}
Hence, the cell constant is 0.219cm−10.219c{m^{ - 1}}.

Note:
Conductivity is measured by determining the amount of current that can be carried between two electrodes by a known amount of liquid. To determine the amount of current that will flow through a “known amount of liquid”, the volume between the two electrodes must be exact.