The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of 0.`88cm^(-1)`. The value of equivalent conductance of solution is

Options

A. 400 mho `cm^(2) g eq^(-1)`
B. 295 mho `cm^(2) g eq^(-1)`
C. 419 mho `cm^(2) g eq^(-1)`
D. 425 mho `cm^(2) g eq^(-1)`

Explanation

To find the equivalent conductance of a 0.01 N solution of an electrolyte given the resistance and cell constant, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance (R) = 220 ohms
- Cell constant (k) = 0.88 cm⁻¹
- Normality (N) = 0.01 N

2. **Calculate Conductance (G):**
Conductance (G) is the reciprocal of resistance (R):
\[ G = \frac{1}{R} = \frac{1}{220} \text{ S} \]

3. **Calculate Specific Conductance (κ):**
The specific conductance (κ) can be calculated using the formula:
\[ κ = G \times k \]
Substituting the values:
\[ κ = \left(\frac{1}{220}\right) \times 0.88 \]

4. **Calculate κ:**
\[ κ = \frac{0.88}{220} \approx 0.0040 \text{ S cm}^{-1} \]

5. **Calculate Equivalent Conductance (λ):**
The formula for equivalent conductance (λ) is:
\[ λ = \frac{κ \times 1000}{N} \]
Substituting the values:
\[ λ = \frac{0.0040 \times 1000}{0.01} \]

6. **Calculate λ:**
\[ λ = \frac{4.0}{0.01} = 400 \text{ S cm}^{2} \text{ eq}^{-1} \]

### Final Answer:
The equivalent conductance of the solution is **400 S cm² eq⁻¹**.