JEE MAIN - Chemistry (2021 - 17th March Evening Shift - No. 13)
A KCl solution of conductivity 0.14 S m$$-$$1 shows a resistance of 4.19$$\Omega$$ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03$$\Omega$$. The conductivity of the HCl solution is ____________ $$\times$$ 10$$-$$2 S m$$-$$1. (Round off to the Nearest Integer).
Answer
57
Explanation
For KCl solution,
$$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$$
= 0.58
For HCl solution,
$$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$$
$$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.03}} = 0.56 = 56 \times {10^{ - 2}}$$ Sm$$-$$1
$$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$$
= 0.58
For HCl solution,
$$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$$
$$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.03}} = 0.56 = 56 \times {10^{ - 2}}$$ Sm$$-$$1
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