Q45 / 150 · MHT CET 2024 (03 May Shift 2) · Chemistry

What is the cell constant, if two platinum electrodes in conductivity cell are separated by 0.92 cm and area of cross section is $1.2 \mathrm{~cm}^2$ ?

  • Correct answer
    A
    $0.767 \mathrm{~cm}^{-1}$
  • B
    $0.906 \mathrm{~cm}^{-1}$
  • C
    $1.304 \mathrm{~cm}^{-1}$
  • D
    $1.104 \mathrm{~cm}^{-1}$

Solution

Cell constant $=\frac{l}{\mathrm{a}}=\frac{0.92 \mathrm{~cm}}{1.2 \mathrm{~cm}^2}=0.767 \mathrm{~cm}^{-1}$

Hence, the correct answer is option: 'A'.

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