The conductivity of 0.02 N solution of a cell of KCl at `25^@C" is "2.765 xx 10^(-3) Scm^(-1)`. If the resistance of a cell containing this solution is 4000 ohm, what will be the cell constant?

Options

A. `11.06 cm^(-1)`
B. `1.21 cm^(-1)`
C. `1.19 cm^(-1)`
D. `1.286cm^(-1)`

Explanation

To find the cell constant of the KCl solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Conductivity (κ) = \(2.765 \times 10^{-3} \, \text{S/cm}\)
- Resistance (R) = \(4000 \, \Omega\)

2. **Understand the relationship between resistance, conductivity, and cell constant**:
The relationship can be expressed as:
\[ R = \frac{1}{\kappa} \times \frac{L}{A} \]
where \(L\) is the length of the cell and \(A\) is the cross-sectional area. The term \(\frac{L}{A}\) is defined as the cell constant (K).

3. **Rearranging the formula to find the cell constant**:
From the equation, we can rearrange it to find the cell constant (K):
\[ K = R \times \kappa \]

4. **Substituting the values into the formula**:
Substitute the values of R and κ into the equation:
\[ K = 4000 \, \Omega \times 2.765 \times 10^{-3} \, \text{S/cm} \]

5. **Calculating the cell constant**:
\[ K = 4000 \times 2.765 \times 10^{-3} = 11.06 \, \text{cm}^{-1} \]

### Final Answer:
The cell constant (K) is \(11.06 \, \text{cm}^{-1}\).

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