MHT CET 2024 2nd May Evening Shift | Electrochemistry Question 40 | Chemistry

The resistance of a conductivity cell of 0.1 M KCl solution is 120 ohm and conductivity is $1.64 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$. What is the value of cell constant?

$0.0136 \mathrm{~cm}^{-1}$

$0.0196 \mathrm{~cm}^{-1}$

$0.0618 \mathrm{~cm}^{-1}$

$0.0731 \mathrm{~cm}^{-1}$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

Which of the following changes occurs during the discharging of lead accumulator?

$\mathrm{Pb}_{(\mathrm{s})}$ is reduced

$\mathrm{H}_2 \mathrm{SO}_4$ is consumed

$\mathrm{PbSO}_4$ is consumed

$\mathrm{PbO}_2$ is produced

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

Calculate the quantity of electricity required to liberate 0.1 mole of chlorine gas during electrolysis of molten sodium chloride.

9665 C

19300 C

14500 C

96500 C

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

Which from following expressions is used to calculate $\mathrm{E}_{\text {cell }}$ for the following cell at $25^{\circ} \mathrm{C}$ ?

$$\mathrm{Pb}_{(\mathrm{s})}\left|\mathrm{Pb}_{(\mathrm{1M})}^{++} \| \mathrm{Ag}_{(\mathrm{10M})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}$$

$\mathrm{E}_{\mathrm{cell}}=\left(\mathrm{E}_{\mathrm{cell}}^{\circ}+0.0592\right) \mathrm{V}$

$\mathrm{E}_{\text {cell }}=\left(\mathrm{E}_{\text {cell }}^{\circ}-0.0592\right) \mathrm{V}$

$\mathrm{E}_{\text {cell }}=\left(\mathrm{E}_{\text {cell }}^{\circ}-0.0296\right) \mathrm{V}$

$\mathrm{E}_{\text {cell }}=\left(\mathrm{E}_{\mathrm{cell}}^{\circ}+0.0296\right) \mathrm{V}$

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