The specific conductance of a decinormal KCl solution is $0.0121\, ohm^{-1}\, cm^{-1}$ and the resistance was found to be 56 ohm. Cell constant for the reaction will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $0.6776\, cm^{-1}$
The relationship is:
$\kappa = \frac{1}{R} \times G^*$
where $\frac{1}{R}$ is the conductance ($G$).
Rearranging to solve for the Cell Constant ($G^*$):
$G^* = \kappa \times R$
Calculation
Identify the given values:
- Specific Conductance ($\kappa$) $= 0.0121\ \Omega^{-1}\ \text{cm}^{-1}$
- Resistance ($R$) $= 56\ \Omega$
Substitute the values into the formula:
- $G^* = \left(0.0121\ \Omega^{-1}\ \text{cm}^{-1}\right) \times (56\ \Omega)$
Calculate the result:
- $G^* = 0.6776\ \text{cm}^{-1}$
Conclusion
The cell constant for the reaction will be $\mathbf{0.6776\ \text{cm}^{-1}}$.