A conductivity cell having cell constant `8.76 cm^(-1)` is placed in 0.01 M solution of an electrolyte offered a resistance of 1000 `"ohm"` . What is the condictivity of electrolytic solution ?

Options

A. `8.76 xx 10^(4) "ohm"^(-1) cm^(-1)`
B. `8.76 xx 10^(-4) "ohm "^(-1) cm^(-1)`
C. `8.76 xx 10^(-3) "ohm"^(-1) cm^(-1)`
D. `8.76 xx 10^(-2) "ohm"^(-1) cm^(-1)`

Answer

To find the conductivity of the electrolytic solution, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Cell constant (k) = 8.76 cm⁻¹
- Resistance (R) = 1000 ohms

2. **Understand the Relationship**:
The relationship between resistance (R), conductance (G), and conductivity (κ) is given by:
\[ G = \frac{1}{R} \]
and
\[ G = k \cdot \kappa \]
where:
- G is the conductance,
- k is the cell constant,
- κ is the conductivity.

3. **Calculate Conductance (G)**:
Using the resistance value:
\[ G = \frac{1}{R} = \frac{1}{1000 \, \text{ohms}} = 0.001 \, \text{S} = 1 \times 10^{-3} \, \text{S} \]

4. **Relate Conductance to Conductivity**:
From the relationship \( G = k \cdot \kappa \), we can express conductivity (κ) as:
\[ \kappa = \frac{G}{k} \]

5. **Substitute Values**:
Now, substituting the values of G and k:
\[ \kappa = \frac{1 \times 10^{-3} \, \text{S}}{8.76 \, \text{cm}^{-1}} \]

6. **Calculate Conductivity**:
Performing the calculation:
\[ \kappa = \frac{1 \times 10^{-3}}{8.76} \approx 0.000112 \, \text{S/cm} = 8.76 \times 10^{-4} \, \text{S/cm} \]

7. **Final Result**:
Therefore, the conductivity of the electrolytic solution is:
\[ \kappa \approx 8.76 \times 10^{-3} \, \text{ohm}^{-1} \text{cm}^{-1} \]