JEE MAIN - Chemistry (2021 - 1st September Evening Shift - No. 22)
If the conductivity of mercury at 0$$^\circ$$C is 1.07 $$\times$$ 106 S m$$-$$1 and the resistance of a cell containing mercury is 0.243$$\Omega$$, then the cell constant of the cell is x $$\times$$ 104 m$$-$$1. The value of x is ____________. (Nearest integer)
Answer
26
Explanation
Conductance (G) is reciprocal of resistance (R)
$$R = {1 \over G}$$ or $$G = {1 \over R} = {1 \over {0.243\Omega }} = 4.115{\Omega ^{ - 1}}$$
Relation between conductance (G),
conductivity ($$\kappa $$) and cell constant $$\left( {{l \over A}} \right)$$ is given as
$$\kappa = {{Gl} \over A}$$
$$ \Rightarrow {l \over A} = {\kappa \over G} = {{1.07 \times {{10}^6}S{m^{ - 1}}} \over {4.115{\Omega ^{ - 1}}}} = 26 \times {10^4}{m^{ - 1}} \Rightarrow x = 26$$
$$R = {1 \over G}$$ or $$G = {1 \over R} = {1 \over {0.243\Omega }} = 4.115{\Omega ^{ - 1}}$$
Relation between conductance (G),
conductivity ($$\kappa $$) and cell constant $$\left( {{l \over A}} \right)$$ is given as
$$\kappa = {{Gl} \over A}$$
$$ \Rightarrow {l \over A} = {\kappa \over G} = {{1.07 \times {{10}^6}S{m^{ - 1}}} \over {4.115{\Omega ^{ - 1}}}} = 26 \times {10^4}{m^{ - 1}} \Rightarrow x = 26$$
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