MHT CET202527 Apr 2025Evening ShiftChemistryElectrochemistryActual
Calculate the cell constant of conductivity cell containing 0.01 M AgNO ₃ solution having resistance 1440 and conductivity 0.001262 ⁻¹ ~cm ⁻¹ .
Options
- A1.014 ~cm ⁻¹
- B0.883 ~cm ⁻¹
- C1.817 ~cm ⁻¹
- D1.411 ~cm ⁻¹
Correct answer
A. 1.014 ~cm ⁻¹
Step-by-step solution
The cell constant G^* relates conductivity and resistance R through the equation G^* = R . Substituting the given values = 0.001262 ⁻¹ cm ⁻¹ and R = 1440 yields G^* = (0.001262) (1440) = 1.81728 cm ⁻¹ . Rounding to three decimal places gives 1.817 cm ⁻¹ , which corresponds to option C .