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Electrochemistry question

2022 · 24 Jun · Shift 2 · Q17
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Electrochemistry question

2022 · 24 Jun · Shift 2 · Q17

JEE MainChemistryElectrochemistryNumerical+4 / −1
The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 Ω\OmegaΩ. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 ×\times× 10 −-− 3 S cm −-− 1, then the cell constant of the conductivity cell is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 3 cm −-− 1.
Numerical answer
View written solutionFree

Correct answer: 266

  1. Given data

    • Resistance of the cell:
      R=1750 ΩR = 1750\,\OmegaR=1750Ω
    • Conductivity of 0.01 M KCl0.01\,\text{M KCl}0.01M KCl at 298 K298\,\text{K}298K:
      κ=0.152×10−3 S cm−1\kappa = 0.152 \times 10^{-3}\,\text{S cm}^{-1}κ=0.152×10−3S cm−1

  2. Relation between conductivity, conductance, and cell constant

    Conductance is: G=1RG = \frac{1}{R}G=R1​

    Also, κ=G×cell constant\kappa = G \times \text{cell constant}κ=G×cell constant

    Therefore, cell constant=κR\text{cell constant} = \kappa Rcell constant=κR

  3. Substitute the values

    cell constant=(0.152×10−3)(1750)\text{cell constant} = (0.152 \times 10^{-3})(1750)cell constant=(0.152×10−3)(1750)

  4. Calculate

    First multiply 0.152×17500.152 \times 17500.152×1750: 0.152×1750=2660.152 \times 1750 = 2660.152×1750=266

    Hence, cell constant=266×10−3 cm−1\text{cell constant} = 266 \times 10^{-3}\,\text{cm}^{-1}cell constant=266×10−3cm−1

  5. Required integer

    Since the answer is asked in the form: ‾×10−3 cm−1\underline{\hspace{2cm}} \times 10^{-3}\,\text{cm}^{-1}​×10−3cm−1

    the required integer is: 266\boxed{266}266​

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