Question 65: The conductivity of centimolar solution of KCl at 25^{\circ} \mathrm{C} is 0.0210 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} and the resistance of the cell containing the solution at 25^{\circ} \mathrm{C} is 60 ohm. The value of cell constant is –
(1) 1.26 \mathrm{~cm}^{-1}
(2) 3.34 \mathrm{~cm}^{-1}
(3) 1.34 \mathrm{~cm}^{-1}
(4) 3.28 \mathrm{~cm}^{-1}
Answer: Option (1)
Explanation:
The relation between conductivity \kappa, resistance R,
and cell constant is given by:
\kappa = \frac{\text{Cell constant}}{R}Rearranging the equation:
\text{Cell constant} = \kappa \times RGiven, \kappa = 0.0210 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} and R = 60 \ \mathrm{ohm}
Substituting the values:
\text{Cell constant} = 0.0210 \times 60 \text{Cell constant} = 1.26 \ \mathrm{cm}^{-1}Hence, the correct value of the cell constant is 1.26 \mathrm{~cm}^{-1},
which corresponds to option (1).