A conductivity cell when filled with 0.02 M KCl (conductivity = 0.002768 `Omega^(-1) cm^(-1)`) has a resistance of 457.3 `Omega`. What will be the equivalent conductivity of `0.05 N CaCl_(2)` solution if the same cell filled with this solution has a resistance of 2020?

Explanation

To solve the problem step by step, we will follow the outlined process to find the equivalent conductivity of the 0.05 N CaCl₂ solution.

### Step 1: Understand the given data
We have the following information:
- Conductivity of 0.02 M KCl solution (κ₁) = 0.002768 Ω⁻¹ cm⁻¹
- Resistance of KCl solution (R₁) = 457.3 Ω
- Resistance of CaCl₂ solution (R₂) = 2020 Ω
- Concentration of CaCl₂ solution (C₂) = 0.05 N

### Step 2: Calculate the cell constant (X)
The cell constant (X) can be calculated using the formula:
\[ X = κ₁ \times R₁ \]
Where:
- κ₁ is the conductivity of KCl solution
- R₁ is the resistance of KCl solution

Substituting the values:
\[ X = 0.002768 \, \Omega^{-1} \text{cm}^{-1} \times 457.3 \, \Omega \]
Calculating this gives:
\[ X = 1.266 \, \text{cm}^{-1} \]

### Step 3: Calculate the conductivity (κ₂) of CaCl₂ solution
The conductivity of the CaCl₂ solution can be calculated using the formula:
\[ κ₂ = \frac{X}{R₂} \]
Where:
- R₂ is the resistance of the CaCl₂ solution

Substituting the values:
\[ κ₂ = \frac{1.266 \, \text{cm}^{-1}}{2020 \, \Omega} \]
Calculating this gives:
\[ κ₂ = 0.000627 \, \Omega^{-1} \text{cm}^{-1} \]

### Step 4: Calculate the equivalent conductivity (λ) of CaCl₂ solution
The equivalent conductivity (λ) can be calculated using the formula:
\[ λ = \frac{κ₂ \times 1000}{C₂} \]
Where:
- C₂ is the concentration of the CaCl₂ solution in N

Substituting the values:
\[ λ = \frac{0.000627 \, \Omega^{-1} \text{cm}^{-1} \times 1000}{0.05} \]
Calculating this gives:
\[ λ = 12.54 \, \Omega^{-1} \text{cm}^2 \text{equivalent}^{-1} \]

### Final Answer
The equivalent conductivity of the 0.05 N CaCl₂ solution is:
\[ \lambda = 12.54 \, \Omega^{-1} \text{cm}^2 \text{equivalent}^{-1} \]

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