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 will follow these steps:

### Step 1: Understand the given data
We have the following information:
- Equivalent conductance (Λ) = 200 Ω⁻¹ cm² eq⁻¹
- Normality (N) = 0.2 N
- Cell constant (g) = 2 cm⁻¹

### Step 2: Use the formula for equivalent conductance
The formula for equivalent conductance is given by:
\[ \Lambda = \frac{1000 \kappa}{N} \]
where:
- Λ = equivalent conductance
- κ = conductivity
- N = normality

### Step 3: Rearrange the formula to find conductivity (κ)
Rearranging the formula to find κ gives us:
\[ \kappa = \frac{\Lambda \cdot N}{1000} \]

### Step 4: Substitute the values into the formula
Substituting the known values into the equation:
\[ \kappa = \frac{200 \, \text{Ω}^{-1} \text{cm}^2 \text{eq}^{-1} \times 0.2 \, \text{eq/L}}{1000} \]

### Step 5: Calculate κ
Calculating κ:
\[ \kappa = \frac{200 \times 0.2}{1000} = \frac{40}{1000} = 0.04 \, \text{Ω}^{-1} \text{cm}^{-1} \]

### Step 6: Use the relationship between conductivity, resistance, and cell constant
The relationship between conductivity (κ), resistance (R), and cell constant (g) is given by:
\[ \kappa = \frac{1}{R} \cdot g \]
Rearranging this gives:
\[ R = \frac{g}{\kappa} \]

### Step 7: Substitute the values to find resistance (R)
Substituting the known values:
\[ R = \frac{2 \, \text{cm}^{-1}}{0.04 \, \text{Ω}^{-1} \text{cm}^{-1}} \]

### Step 8: Calculate R
Calculating R:
\[ R = \frac{2}{0.04} = 50 \, \text{Ω} \]

### Conclusion
The resistance of the solution is **50 Ω**.

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