Which of the following expressions for conductivity of solution of an electrolyte is NOT correct?

Conductivity (\(\mathrm{\kappa}\)) is defined as the reciprocal of resistivity (\(\rho\)). Thus:
\(\kappa=\frac{1}{\rho}\)
For a given cell, the relationship between conductivity and conductance (G) is:
\(\kappa=G \times(\text {cell constant})\)
The cell constant is often denoted by \(\frac{l}{A}\) (length of the conducting path over the cross-sectional area).
Checking each option:
Option A: \(\kappa=\frac{1}{\rho}\)
This is the fundamental definition: Conductivity is the inverse of resistivity.
This is correct.
Option B: \(\kappa=G \frac{1}{a}\)
If \(a\) represents the cell constant (which is usually \(\frac{l}{A}\)), the correct formula is:
\(\kappa=G \times a\)
not \(\kappa=G \frac{1}{a}\).
This does not match the standard relationship, making this suspicious.
Likely incorrect.
Option C: \(\kappa=\frac{1}{R} \frac{a}{1}\)
Since \(G=\frac{1}{R}\), this can be rewritten as:
\(\kappa=G \times a\)
This is the correct formula if \(a\) is the cell constant.
This is correct.
Option D: \(\kappa=\Lambda \cdot c\)
Molar conductivity \(\Lambda\) is defined by:
\(\Lambda=\frac{\kappa}{c} \Longrightarrow \kappa=\Lambda \times c\)
This is a standard relation.
This is correct.
Conclusion:
The only formula that does not align with the standard relationships is:
Option B: \(\kappa=G \frac{1}{a}\)
This formula is not correct based on standard definitions.