A conductivity cell has a cell constant of `0.5 cm^(-1)`. This cell when filled with 0.01 M NaCl solution has a resistance of 384 ohms at `25^(@)C`. Calculate the equivalent conductance of the given solution.

Options

A. `130.2 Omega^(-1) cm^(2) (g eq)^(-1)`
B. `137.4 Omega^(-1) cm^(2) (g eq)^(-1)`
C. `154.6 Omega^(-1) cm^(2) (g eq)^(-1)`
D. `169.2 Omega^(-1) cm^(2) (g eq)^(-1)`

Answer

To calculate the equivalent conductance of the given 0.01 M NaCl solution using the provided data, we will follow these steps:

### Step 1: Understand the relationship between conductance, resistance, and cell constant
The conductance (C) of the solution can be calculated using the formula:
\[ C = \frac{1}{R} \]
where \( R \) is the resistance of the solution.

### Step 2: Calculate the conductance of the solution
Given that the resistance \( R = 384 \, \Omega \):
\[ C = \frac{1}{384 \, \Omega} \]
Calculating this gives:
\[ C \approx 0.002604 \, \text{S} \, (\text{Siemens}) \]

### Step 3: Use the cell constant to find specific conductance (κ)
The specific conductance (κ) can be calculated using the formula:
\[ \kappa = C \times \text{cell constant} \]
Given that the cell constant is \( 0.5 \, \text{cm}^{-1} \):
\[ \kappa = 0.002604 \, \text{S} \times 0.5 \, \text{cm}^{-1} \]
Calculating this gives:
\[ \kappa \approx 0.001302 \, \text{S/cm} \]

### Step 4: Convert molarity to normality
For NaCl, the normality (N) is equal to the molarity (M) since it dissociates into one Na⁺ and one Cl⁻ ion:
\[ N = 0.01 \, \text{M} \]

### Step 5: Calculate the equivalent conductance (Λ)
The equivalent conductance (Λ) is given by the formula:
\[ \Lambda = \frac{\kappa \times 1000}{N} \]
Substituting the values we have:
\[ \Lambda = \frac{0.001302 \, \text{S/cm} \times 1000}{0.01} \]
Calculating this gives:
\[ \Lambda = 130.2 \, \text{S cm}^2/\text{equivalent} \]

### Final Answer
The equivalent conductance of the 0.01 M NaCl solution is:
\[ \Lambda \approx 130.2 \, \text{S cm}^2/\text{equivalent} \]