A weak electrolyte having the limiting equivalent conductance of `400 S cm^(2) g "equivalent"^(-1)` at 298 K is 2% ionized in its 0.1N solution. The resistance of this solution (in ohms) in an electrolytic cell of cell constant `0.4 cm^(-1)` at this temperature is-

Options

A. 200
B. 300
C. 400
D. 500

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Equivalent Conductance (λ) of the solution
Given that the limiting equivalent conductance (λ₀) is 400 S cm² g equivalent⁻¹ and the degree of ionization (α) is 2% (or 0.02), we can calculate the equivalent conductance (λ) of the solution using the formula:

\[ \lambda = \alpha \cdot \lambda_0 \]

Substituting the values:

\[ \lambda = 0.02 \cdot 400 = 8 \, \text{S cm}^2 \text{g equivalent}^{-1} \]

### Step 2: Calculate the Conductivity (κ) of the solution
The conductivity (κ) can be calculated using the formula:

\[ \kappa = \frac{\lambda \cdot N}{1000} \]

where N is the normality of the solution. Given that the normality (N) is 0.1 N, we substitute the values:

\[ \kappa = \frac{8 \cdot 0.1}{1000} = \frac{0.8}{1000} = 0.0008 \, \text{S cm}^{-1} \]

### Step 3: Calculate the Resistance (R) of the solution
The resistance (R) can be calculated using the formula:

\[ R = \frac{l}{\kappa \cdot A} \]

However, we can also use the relationship between resistance, conductivity, and cell constant (K):

\[ R = \frac{1}{\kappa \cdot K} \]

where K is the cell constant given as 0.4 cm⁻¹. Substituting the values:

\[ R = \frac{1}{0.0008 \cdot 0.4} \]

Calculating this gives:

\[ R = \frac{1}{0.00032} = 3125 \, \text{ohms} \]

### Final Answer
The resistance of the solution in the electrolytic cell is **3125 ohms**.

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