The equivalent conductance of a 0.2 n solution of an electrolyte was found to be `200Omega^(-1) cm^(2)eq^(-1)` . The cell constant of the cell is `2 cm^(-1)` . The resistance of the solution is

Options

A. `50Omega`
B. `400Omega`
C. `100Omega`
D. None of these

Answer

To find the resistance of the solution, we can use the relationship between equivalent conductance (Λ), specific conductivity (κ), and resistance (R). Here’s a step-by-step solution:

### Step 1: Understand the relationship
The equivalent conductance (Λ) is related to the specific conductivity (κ) and normality (N) by the formula:
\[ \Lambda = \frac{\kappa \times 1000}{N} \]
Where:
- Λ = equivalent conductance in Ω⁻¹ cm² eq⁻¹
- κ = specific conductivity in Ω⁻¹ cm⁻¹
- N = normality in eq/L

### Step 2: Rearrange the formula to find κ
We can rearrange the formula to solve for κ:
\[ \kappa = \frac{\Lambda \times N}{1000} \]

### Step 3: Substitute the known values
Given:
- Λ = 200 Ω⁻¹ cm² eq⁻¹
- N = 0.2 N

Substituting the values:
\[ \kappa = \frac{200 \times 0.2}{1000} = \frac{40}{1000} = 0.04 \, \text{Ω}^{-1} \text{cm}^{-1} \]

### Step 4: Use the cell constant to find resistance
The specific conductivity (κ) is also related to resistance (R) and the cell constant (K) by the formula:
\[ \kappa = \frac{K}{R} \]
Where:
- K = cell constant (L/A)

Rearranging this gives us:
\[ R = \frac{K}{\kappa} \]

### Step 5: Substitute the known values
Given:
- K = 2 cm⁻¹
- κ = 0.04 Ω⁻¹ cm⁻¹

Substituting the values:
\[ R = \frac{2}{0.04} = 50 \, \text{Ω} \]

### Final Answer
The resistance of the solution is **50 Ω**.

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