Read the passage given below and answer the question:
The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations and at different temperatures. consider the resistance of a conductivity cell filled with 0.1 M KCl solution is 200 Ohm. if the resistance of the same cell when filled with 0.02 M KCl solution is 420 Ohm.
(Conductivity of 0.1 M KCl solution is 1.29 S `m^(-1)`).
Q. SI unit for conductivity of a solution is

Options

A. S `m^(-1)`
B. `Sm^(2)mol^(-1)`
C. `mol" "cm^(-3)`
D. `S" "cm^(2)mol^(-1)`

Explanation

To determine the SI unit for the conductivity of a solution, we can follow these steps:

### Step 1: Understand the Definition of Conductivity
Conductivity (\( \kappa \)) is defined as the ability of a solution to conduct electric current. It is related to the conductance of the solution and the cell constant.

### Step 2: Identify the Relationship
The relationship between conductivity (\( \kappa \)), conductance (\( G \)), and cell constant (\( K \)) can be expressed as:
\[ \kappa = G \times K \]
Where:
- \( G \) is the conductance measured in Siemens (S), which is equivalent to Ohm\(^{-1}\).
- \( K \) is the cell constant, measured in meters\(^{-1}\) (m\(^{-1}\)).

### Step 3: Determine the Units
From the relationship:
- Conductance \( G \) has units of Siemens (S), which is equivalent to Ohm\(^{-1}\).
- The cell constant \( K \) has units of meters\(^{-1}\) (m\(^{-1}\)).

Thus, the units for conductivity can be derived as follows:
\[ \text{Units of } \kappa = \text{Units of } G \times \text{Units of } K = \text{S} \times \text{m}^{-1} \]
This simplifies to:
\[ \text{Units of } \kappa = \text{S m}^{-1} \]

### Step 4: Conclusion
The SI unit for conductivity of a solution is therefore:
\[ \text{S m}^{-1} \]