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 (also known as conductivity) of a 0.0025 M solution of K₂SO₄ with a resistance of 326 ohms and a cell constant of 4, we can follow these steps:

### Step 1: Understand the relationship between resistance and conductivity
The specific conductance (κ) is related to resistance (R) and the cell constant (k) by the formula:
\[ \kappa = \frac{k}{R} \]
where:
- \( \kappa \) is the specific conductance (in S/m),
- \( k \) is the cell constant (in m⁻),
- \( R \) is the resistance (in ohms).

### Step 2: Substitute the given values into the formula
From the problem, we have:
- Resistance \( R = 326 \, \Omega \)
- Cell constant \( k = 4 \)

Now, substituting these values into the formula:
\[ \kappa = \frac{4}{326} \]

### Step 3: Calculate the specific conductance
Now, perform the calculation:
\[ \kappa = \frac{4}{326} \approx 0.01227 \, \text{S/m} \]

### Step 4: Convert to scientific notation
To express this value in scientific notation:
\[ 0.01227 = 1.227 \times 10^{-2} \, \text{S/m} \]
Rounding to two significant figures gives:
\[ \kappa \approx 1.20 \times 10^{-2} \, \text{S/m} \]

### Conclusion
Thus, the specific conductance of the solution is:
\[ \kappa = 1.20 \times 10^{-2} \, \text{S/m} \]

### Final Answer
The correct option is option number 4: \( 1.20 \times 10^{-2} \).

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