For an electrolyte solution of `0.05 mol L^(-1)` , the conductivity has been found to be `0.0110S cm^(-1)`. The molar conductivity is

Options

A. `0.05S cm^(2)mol ^(-1)`
B. `550 S cm^(2) mol ^(-1)`
C. `0.22 S cm^(2) mol ^(-1)`
D. `220 S cm^(2) mol ^(-1)`

Explanation

To find the molar conductivity of the electrolyte solution, we can use the formula:

\[ \Lambda_m = \frac{K \times 1000}{C} \]

Where:
- \(\Lambda_m\) is the molar conductivity (in S cm² mol⁻¹),
- \(K\) is the conductivity of the solution (in S cm⁻¹),
- \(C\) is the molarity of the solution (in mol L⁻¹).

Given:
- \(K = 0.0110 \, \text{S cm}^{-1}\)
- \(C = 0.05 \, \text{mol L}^{-1}\)

Now, we can substitute the values into the formula:

1. **Substituting the values:**
\[ \Lambda_m = \frac{0.0110 \, \text{S cm}^{-1} \times 1000}{0.05 \, \text{mol L}^{-1}} \]

2. **Calculating the numerator:**
\[ 0.0110 \times 1000 = 11.0 \, \text{S cm}^{-1} \]

3. **Now, divide by the molarity:**
\[ \Lambda_m = \frac{11.0 \, \text{S cm}^{-1}}{0.05 \, \text{mol L}^{-1}} = 220 \, \text{S cm}^2 \text{mol}^{-1} \]

Thus, the molar conductivity of the electrolyte solution is:

\[ \Lambda_m = 220 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer:
The molar conductivity is \(220 \, \text{S cm}^2 \text{mol}^{-1}\).

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