A sample of water from a large swimming pool has a resistance of `10000Omega` at `25^(@)C` when placed in a certain conductace cell. When filled with `0.02M KCI` solution, the cell has a resistance of `100 Omega` at `25^(@)C, 585 gm` of `NaCI` were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of `8000 Omega`.
Given: Molar conductance of `NaCI` at that concentration is `125 Omega^(-1) cm^(2) mol^(-1)` and molar conductivity of `KCI` at `0.02 M` is `200W^(-1) cm^(2) mol^(-1)`.
Cell constant (in `cm^(-1))` of conductane cell is:

Options

A. 4
B. 0.4
C. `4 xx 10^(-2)`
D. `4 xx 10^(-5)`

Explanation

To find the cell constant of the conductance cell, we can follow these steps:

### Step 1: Calculate the Specific Conductivity of KCl
Given the molar conductance (Λ) of KCl at 0.02 M is 200 Ω⁻¹ cm² mol⁻¹. The formula for molar conductance is:

\[ \Lambda = \frac{\kappa \cdot 1000}{C} \]

Where:
- \(\Lambda\) = Molar conductance
- \(\kappa\) = Specific conductivity
- \(C\) = Concentration in mol/L

Rearranging the formula to find specific conductivity (\(\kappa\)):

\[ \kappa = \frac{\Lambda \cdot C}{1000} \]

Substituting the values:

\[ \kappa = \frac{200 \cdot 0.02}{1000} = \frac{4}{1000} = 0.004 \, \text{S/cm} = 4 \times 10^{-3} \, \text{S/cm} \]

### Step 2: Relate Specific Conductivity to Resistance
The specific conductivity is also related to the resistance (R) and the cell constant (G* or L/A) by the formula:

\[ \kappa = \frac{G^*}{R} \]

Rearranging this gives:

\[ G^* = \kappa \cdot R \]

### Step 3: Calculate the Cell Constant for KCl
Now, substituting the specific conductivity (\(\kappa\)) and the resistance for KCl (R = 100 Ω):

\[ G^* = (4 \times 10^{-3}) \cdot 100 = 0.4 \, \text{cm}^{-1} \]

### Final Answer
The cell constant (G*) of the conductance cell is:

\[ \boxed{0.4 \, \text{cm}^{-1}} \]

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