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

$0.6776\, cm^{-1}$

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 $\mathbf{0.6776\ \text{cm}^{-1}}$.