The resistance of a cell (cell constant `= 1.1 cm^(-1)`) containing `(N)/(50)` KCl was found to be 400 ohms. Find the equivalent conductivity of KCl at this dilution.

Explanation

To find the equivalent conductivity of KCl at the given dilution, we can follow these steps:

### Step 1: Understand the given data
- Cell constant (k) = 1.1 cm⁻¹
- Resistance (R) = 400 ohms
- Normality (N) = N/50

### Step 2: Calculate the conductivity (K)
The conductivity (K) can be calculated using the formula:
\[ K = \frac{\text{Cell constant}}{\text{Resistance}} \]
Substituting the values:
\[ K = \frac{1.1 \, \text{cm}^{-1}}{400 \, \Omega} \]

### Step 3: Perform the calculation for K
Calculating the above expression:
\[ K = \frac{1.1}{400} = 0.00275 \, \text{S/cm} \]

### Step 4: Calculate the volume (V) of KCl solution
Since the normality is given as N/50, we can find the volume of the solution. Normality (N) is defined as the number of equivalents of solute per liter of solution. Therefore, for N/50:
\[ \text{Volume (V)} = \frac{50 \, \text{equivalents}}{1 \, \text{N}} \times 1000 \, \text{mL} = 1000 \, \text{mL} \]

### Step 5: Calculate the equivalent conductivity (λ)
The equivalent conductivity (λ) can be calculated using the formula:
\[ \lambda = K \times V \]
Substituting the values:
\[ \lambda = 0.00275 \, \text{S/cm} \times 1000 \, \text{mL} \]

### Step 6: Perform the calculation for λ
Calculating the above expression:
\[ \lambda = 0.00275 \times 1000 = 2.75 \, \text{S cm}^2/\text{equivalent} \]

### Final Answer
The equivalent conductivity of KCl at this dilution is:
\[ \lambda = 2.75 \, \text{S cm}^2/\text{equivalent} \]

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