
The unit of the cell constant is :
A. $oh{m^{ - 1}}$
B. $ohm - cm$
C. $c{m^{ - 1}}$
D. $oh{m^{ - 1}}c{m^2}e{q^{ - 1}}$
Answer
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Hint: Cell constant denoted by ${G^*}$ is defined as the ratio of the distance between the electrodes to the electrode area . The value of cell constant differs from cell to cell .
Complete step by step answer:
As we know conductivity is given by
$\kappa = G \times \dfrac{l}{a}$
where , G is conductance of solution
l is length of the conductor
and a is the area of cross - section of the conductor .
The measurement of l and a is difficult . However , for a particular cell , $\dfrac{l}{a}$ is constant and this constant is known as cell constant .
Hence , we find out that ${G^*} = \dfrac{l}{a}$ where l is length and a is an area of cross-section of conductor.
Now , since unit of l is $cm$ and unit of a is $c{m^2}$ , we have
${G^*} = \dfrac{{cm}}{{c{m^2}}}$
$ \Rightarrow {G^*} = c{m^{ - 1}}$
Therefore , the unit of cell constant is $c{m^{ - 1}}$ .
So, the correct answer is “Option C”.
Additional Information: If we want to find out the cell constant of any particular cell it can be found out by measuring the conductance of a solution whose conductivity is known . For this purpose we use specific conductivities of $KCl$ solutions which are known accurately at different concentrations and temperatures .
Once the value of cell constant is known , the conductivity of the given solution can be determined by measuring its conductance and multiplying the value with the cell constant .
Note: The value of cell constant remains constant but the value of conductance decreases with increase in temperature due to movement of kernels ( nucleus with inner electrons ) as there is hindrance in the flow of electrons .
Complete step by step answer:
As we know conductivity is given by
$\kappa = G \times \dfrac{l}{a}$
where , G is conductance of solution
l is length of the conductor
and a is the area of cross - section of the conductor .
The measurement of l and a is difficult . However , for a particular cell , $\dfrac{l}{a}$ is constant and this constant is known as cell constant .
Hence , we find out that ${G^*} = \dfrac{l}{a}$ where l is length and a is an area of cross-section of conductor.
Now , since unit of l is $cm$ and unit of a is $c{m^2}$ , we have
${G^*} = \dfrac{{cm}}{{c{m^2}}}$
$ \Rightarrow {G^*} = c{m^{ - 1}}$
Therefore , the unit of cell constant is $c{m^{ - 1}}$ .
So, the correct answer is “Option C”.
Additional Information: If we want to find out the cell constant of any particular cell it can be found out by measuring the conductance of a solution whose conductivity is known . For this purpose we use specific conductivities of $KCl$ solutions which are known accurately at different concentrations and temperatures .
Once the value of cell constant is known , the conductivity of the given solution can be determined by measuring its conductance and multiplying the value with the cell constant .
Note: The value of cell constant remains constant but the value of conductance decreases with increase in temperature due to movement of kernels ( nucleus with inner electrons ) as there is hindrance in the flow of electrons .
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