The resistance of `0.01` N solution of an electrolyte waw found to be 210 ohm at 298 K, when a conductivity cell with cell constant `0.66 cm^(-1)` is used. The equivalent conductance of solution is:

Options

A. `314.28` mho `cm^(2)eq^(-1)`
B. `3.14` mho `cm^(2)eq^(-1)`
C. `31.428` mho `cm^(2)eq^(-1)`
D. `0.314` mho `cm^(2)eq^(-1)`

Answer

To find the equivalent conductance of a 0.01 N solution of an electrolyte, we can follow these steps:

### Step 1: Calculate the Specific Conductivity (κ)
The specific conductivity (κ) can be calculated using the formula:
\[ \kappa = \frac{R}{K} \]
Where:
- \( R \) is the resistance of the solution (210 ohms),
- \( K \) is the cell constant (0.66 cm\(^{-1}\)).

Using the formula:
\[ \kappa = \frac{1}{R \cdot K} \]
Substituting the values:
\[ \kappa = \frac{1}{210 \cdot 0.66} = \frac{0.66}{210} \approx 0.00314 \, \text{S/cm} \]

### Step 2: Calculate the Equivalent Conductance (Λ)
The equivalent conductance (Λ) can be calculated using the formula:
\[ \Lambda = \frac{\kappa \times 1000}{N} \]
Where:
- \( \kappa \) is the specific conductivity (0.00314 S/cm),
- \( N \) is the normality of the solution (0.01 N).

Substituting the values:
\[ \Lambda = \frac{0.00314 \times 1000}{0.01} = \frac{3.14}{0.01} = 314 \, \text{ohm}^{-1} \text{cm}^2 \text{equivalent}^{-1} \]

### Final Answer
The equivalent conductance of the solution is:
\[ \Lambda \approx 314 \, \text{ohm}^{-1} \text{cm}^2 \text{equivalent}^{-1} \]

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