One normal salt solution surrounding two platinum electrodes, 2.1 cm apart and `6.3 cm^(2)` in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity solution.

Answer

To calculate the equivalent conductivity of the salt solution, we will follow these steps:

### Step 1: Identify the given values
- Distance between electrodes (l) = 2.1 cm
- Area of electrodes (A) = 6.3 cm²
- Resistance (R) = 50 ohms

### Step 2: Convert the distance to meters
Since conductivity is usually expressed in SI units, we convert the distance from centimeters to meters.
- \( l = 2.1 \, \text{cm} = 0.021 \, \text{m} \)

### Step 3: Calculate the specific conductance (k)
The formula for specific conductance (k) is given by:
\[ k = \frac{l}{A} \times \frac{1}{R} \]
Substituting the values:
\[ k = \frac{0.021 \, \text{m}}{6.3 \, \text{cm}^2} \times \frac{1}{50 \, \Omega} \]
Note: Convert area from cm² to m²:
\[ A = 6.3 \, \text{cm}^2 = 6.3 \times 10^{-4} \, \text{m}^2 \]
Now substituting:
\[ k = \frac{0.021}{6.3 \times 10^{-4}} \times \frac{1}{50} \]

### Step 4: Calculate the specific conductance value
Calculating the above expression:
\[ k = \frac{0.021}{0.00063} \times \frac{1}{50} \]
\[ k = 33.33 \times \frac{1}{50} = 0.6666 \, \text{S/m} \]

### Step 5: Convert specific conductance to equivalent conductivity
The equivalent conductivity (Λ) is given by:
\[ \Lambda = k \times V \]
Where V is the volume of the solution. Since we have a 1 normal solution, we assume the volume (V) to be 1 liter (1000 mL).
\[ \Lambda = 0.6666 \, \text{S/m} \times 1 \, \text{L} = 0.6666 \, \text{S m}^2/\text{mol} \]

### Step 6: Convert to appropriate units
Convert S/m to ohm⁻¹ cm⁻¹:
\[ 0.6666 \, \text{S/m} = 0.06666 \, \text{ohm}^{-1} \, \text{cm}^{-1} \]
Thus,
\[ \Lambda = 0.06666 \, \text{ohm}^{-1} \, \text{cm}^{-1} \times 1000 = 66.66 \, \text{ohm}^{-1} \, \text{cm}^{-1} \]

### Final Answer
The equivalent conductivity of the solution is approximately:
\[ \Lambda \approx 66.66 \, \text{ohm}^{-1} \, \text{cm}^{-1} \]