0.1 mole per litre solution present in conductivity cell where electrode of 100 `cm^(2)` area placed at 1 cm and resistance observe is `5xx10^(3)` ohm, what is molar conductivity of solution?

Options

A. `5xx10^(2)S cm^(2) "mole"^(-1)`
B. `10^(4)S cm^(2) "mole"^(-1)`
C. `200 S cm^(2)"mole"^(-1)`
D. `0.02 S cm^(2) "mole"^(-1)`

Explanation

To find the molar conductivity of the solution, we can follow these steps:

### Step 1: Understand the given data
We have:
- Molarity (C) = 0.1 moles per liter
- Area of electrodes (A) = 100 cm²
- Distance between electrodes (L) = 1 cm
- Resistance (R) = 5 × 10³ ohms

### Step 2: Calculate the cell constant (k)
The cell constant (k) is given by the formula:
\[ k = \frac{L}{A} \]
Substituting the values:
\[ k = \frac{1 \, \text{cm}}{100 \, \text{cm}^2} = 0.01 \, \text{cm}^{-1} \]

### Step 3: Calculate specific conductivity (κ)
Specific conductivity (κ) can be calculated using the formula:
\[ \kappa = \frac{1}{R} \times k \]
Substituting the values:
\[ \kappa = \frac{1}{5 \times 10^3 \, \Omega} \times 0.01 \, \text{cm}^{-1} \]
\[ \kappa = \frac{0.01}{5 \times 10^3} \]
\[ \kappa = 2 \times 10^{-6} \, \text{S cm}^{-1} \]

### Step 4: Calculate molar conductivity (Λ)
Molar conductivity (Λ) is calculated using the formula:
\[ \Lambda = \kappa \times \frac{1000}{C} \]
Where C is the molarity in moles per liter. Substituting the values:
\[ \Lambda = 2 \times 10^{-6} \, \text{S cm}^{-1} \times \frac{1000}{0.1} \]
\[ \Lambda = 2 \times 10^{-6} \times 10^{4} \]
\[ \Lambda = 2 \times 10^{-2} \, \text{S cm}^{-1} \]
\[ \Lambda = 0.02 \, \text{S cm}^{-1} \]

### Conclusion
The molar conductivity of the solution is:
\[ \Lambda = 0.02 \, \text{S cm}^{-1} \]

### Answer
The correct option is 0.02 S cm² mol⁻¹.

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