At 298K the resistance of a 0.5N NaOH solution is 35.0 ohm. The cell constant is 0.503 `cm^(-1)` the electrical conductivity of the solution is

Options

A. `1.437xx10^(-2)ohm ^(-1)cm^(-1)`
B. `1.473ohm^(-1)cm^(-1)`
C. `1.06ohm^(-1)cm^(-1)`
D. `3.5ohm^(-1)cm^(-1)`.

Explanation

To find the electrical conductivity of the 0.5N NaOH solution, we can use the relationship between conductivity (κ), resistance (R), and cell constant (G). The formula we will use is:

\[ \kappa = G \cdot \frac{1}{R} \]

where:
- \( \kappa \) is the electrical conductivity,
- \( G \) is the cell constant,
- \( R \) is the resistance.

### Step-by-Step Solution:

1. **Identify the given values**:
- Resistance, \( R = 35.0 \, \Omega \)
- Cell constant, \( G = 0.503 \, \text{cm}^{-1} \)

2. **Calculate the conductance (G)**:
Conductance is the reciprocal of resistance:
\[ G = \frac{1}{R} = \frac{1}{35.0 \, \Omega} \]

3. **Calculate the conductance**:
\[ G = \frac{1}{35.0} \approx 0.02857 \, \text{S} \, \text{(Siemens)} \]

4. **Calculate the electrical conductivity (κ)**:
Now, substitute the values of conductance and cell constant into the conductivity formula:
\[ \kappa = G \cdot \frac{1}{R} = 0.503 \, \text{cm}^{-1} \cdot 0.02857 \, \text{S} \]

5. **Perform the calculation**:
\[ \kappa = 0.503 \cdot 0.02857 \approx 0.01437 \, \text{S/cm} \]

6. **Convert to appropriate units**:
Since \( 1 \, \text{S/cm} = 10^2 \, \text{mS/cm} \), we can express it as:
\[ \kappa \approx 1.437 \times 10^{-2} \, \text{S/cm} \]

### Final Answer:
The electrical conductivity of the 0.5N NaOH solution is approximately \( 1.437 \times 10^{-2} \, \text{S/cm} \).