The molar conductivity of 0.04 M solution of `MgCl_2` is 200 `Scm^2mol^(-1)` at 298 k. A cell with electrodes that are 2.0`cm^2` in surface area and 0.50cm apart is filled with `MgCl_2` solution.
Conductance of `MgCl_2` solution is :

Options

A. `8xx10^(-3)`S
B. 32S
C. 0.032S
D. None of these

Answer

To find the conductance of the `MgCl_2` solution, we can follow these steps:

### Step 1: Calculate the Specific Conductance (K)
The relationship between molar conductivity (Λ) and specific conductance (K) is given by the formula:

\[ \Lambda = \frac{K \times 1000}{M} \]

Where:
- \(\Lambda\) = Molar conductivity = 200 S cm² mol⁻¹
- \(M\) = Molarity = 0.04 mol/L

Rearranging the formula to find K:

\[ K = \frac{\Lambda \times M}{1000} \]

Substituting the values:

\[ K = \frac{200 \, \text{S cm}^2 \text{mol}^{-1} \times 0.04 \, \text{mol/L}}{1000} \]

Calculating:

\[ K = \frac{8}{1000} = 0.008 \, \text{S cm}^{-1} \]

### Step 2: Use the Conductance Formula
The conductance (G) of the solution can be calculated using the formula:

\[ G = \frac{K \times A}{L} \]

Where:
- \(A\) = Area of the electrodes = 2.0 cm²
- \(L\) = Distance between the electrodes = 0.50 cm

Substituting the values:

\[ G = \frac{0.008 \, \text{S cm}^{-1} \times 2.0 \, \text{cm}^2}{0.50 \, \text{cm}} \]

Calculating:

\[ G = \frac{0.016}{0.50} = 0.032 \, \text{S} \]

### Final Answer
The conductance of the `MgCl_2` solution is **0.032 S**.

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