The resistance of 0.0025M solution of `K_(2)SO_(4)` is 326ohm. The specific conductance of the solution, if cell constant is 4.

Options

A. `4.997 xx 10^(-4)`
B. `5.99 xx 10^(-7)`
C. `6.99 xx 10^(-4)`
D. `1.20 xx 10^(-2)`

Answer

To find the specific conductance (κ) of a 0.0025 M solution of K₂SO₄ given its resistance (R) and the cell constant (k), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
The specific conductance (κ) is related to the resistance (R) and the cell constant (k) by the formula:
\[ κ = \frac{k}{R} \]

2. **Identify Given Values**:
- Resistance (R) = 326 ohms
- Cell constant (k) = 4

3. **Substitute the Values into the Formula**:
Using the values we have:
\[ κ = \frac{4}{326} \]

4. **Calculate Specific Conductance**:
Now, perform the division:
\[ κ = \frac{4}{326} \approx 0.01227 \, \text{S/m} \]

5. **Convert to Scientific Notation**:
To express this in scientific notation:
\[ κ \approx 1.227 \times 10^{-2} \, \text{S/m} \]

6. **Final Answer**:
Thus, the specific conductance of the solution is:
\[ κ \approx 1.20 \times 10^{-2} \, \text{S/m} \]