If cell constant is `0.40 cm^(-1)`, the conductivity of 0.015 M NaCl solution having R = 1850 ohm is equal to

Options

A. `1.08xx10^(-4)S cm^(-1)`
B. `4.32xx10^(-4)S cm^(-1)`
C. `2.16xx10^(-4)S cm^(-1)`
D. `5.04xx10^(-5)Scm^(-1)`

Explanation

To find the conductivity of the 0.015 M NaCl solution, we can use the formula:

\[ \text{Conductivity} (\kappa) = \frac{\text{Cell Constant} (k)}{\text{Resistance} (R)} \]

### Step-by-Step Solution:

1. **Identify the given values**:
- Cell constant, \( k = 0.40 \, \text{cm}^{-1} \)
- Resistance, \( R = 1850 \, \Omega \)

2. **Plug in the values into the formula**:
\[ \kappa = \frac{0.40 \, \text{cm}^{-1}}{1850 \, \Omega} \]

3. **Perform the division**:
- First, calculate \( \frac{0.40}{1850} \):
\[ \kappa = \frac{0.40}{1850} = 0.000216216 \, \text{cm}^{-1} \, \Omega^{-1} \]

4. **Convert the result into scientific notation**:
\[ \kappa = 2.16 \times 10^{-4} \, \text{cm}^{-1} \, \Omega^{-1} \]

5. **Convert the units**:
- Since \( \Omega^{-1} \) is equivalent to S (Siemens), we can write:
\[ \kappa = 2.16 \times 10^{-4} \, \text{cm}^{-1} \, \text{S} \]

### Final Answer:
The conductivity of the 0.015 M NaCl solution is:
\[ \kappa = 2.16 \times 10^{-4} \, \text{cm}^{-1} \, \text{S} \]