Given `l//a=0.5 cm^(-1), R= 50 "ohm", N= 1.0.` The equivalent conductance of the electrolytic cell is .

Options

A. `10Omega^(-1) cm^(2) equ iv ^(-1)`
B. `(20Omega^(-1) cm^(2) g equ iv ^(-1)`
C. `300Omega^(-1)cm^(2) g equ iv ^(-1)`
D. ` 100Omega^(-1)cm ^(2) g equiv ^(-1)`

Explanation

To find the equivalent conductance of the electrolytic cell, we can follow these steps:

### Step 1: Understand the given values
We have the following values:
- Cell constant (l/A) = 0.5 cm^(-1)
- Resistance (R) = 50 ohms
- Normality (N) = 1.0

### Step 2: Calculate specific conductivity (κ)
The formula for specific conductivity (κ) is given by:
\[ \kappa = \text{Cell Constant} \times \frac{1}{\text{Resistance}} \]
Substituting the given values:
\[ \kappa = 0.5 \, \text{cm}^{-1} \times \frac{1}{50 \, \text{ohm}} = \frac{0.5}{50} = 0.01 \, \text{S/cm} = 10^{-2} \, \text{S/cm} \]

### Step 3: Calculate equivalent conductance (Λ)
The formula for equivalent conductance (Λ) is:
\[ \Lambda = \frac{\kappa \times 1000}{N} \]
Substituting the values we have:
\[ \Lambda = \frac{10^{-2} \, \text{S/cm} \times 1000}{1} = 10 \, \text{S cm}^{-1} \text{equivalent}^{-1} \]

### Conclusion
The equivalent conductance of the electrolytic cell is:
\[ \Lambda = 10 \, \text{S cm}^{-1} \text{equivalent}^{-1} \]

### Final Answer
The equivalent conductance of the electrolytic cell is **10 S cm⁻¹ equivalent⁻¹**.

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