JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Redox Reactions and Electrochemistry

A KCl solution of conductivity

0.14 Sm^{- 1}

shows a resistance of

4.19Ω

in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to

1.03Ω

. The conductivity of the HCl solution is ______

\times 1 0^{- 2} S m^{- 1}

.
(Round off to the nearest integer).

Answer: 57

Solution

The cell constant depends only on the geometry of the electrodes, so it is the same for both solutions and can be found from the standard potassium chloride measurement:

G^{*} = κ R = 0.14 \times 4.19 = 0.5866 m^{- 1}

Applying this cell constant to the hydrochloric acid solution in the same cell,

κ = \frac{G ^{*}}{R} = \frac{0.5866}{1.03} = 0.5695 S m^{- 1}

The conductivity of the hydrochloric acid solution is

57 \times 1 0^{- 2} S m^{- 1}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Redox Reactions and Electrochemistry

Solution

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