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)`.
Volume (in Litres) of water in the pool is:

Options

A. `1.25 xx 10^(-5)`
B. `1250`
C. `12500`
D. None of these

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript and perform the necessary calculations.

### Step 1: Calculate the Cell Constant for KCl
Given:
- Molar conductivity of KCl (\( \Lambda_{KCl} \)) = 200 \( \Omega^{-1} \, cm^2 \, mol^{-1} \)
- Concentration of KCl = 0.02 M

The relationship between molar conductivity (\( \Lambda \)), specific conductivity (\( \kappa \)), and concentration (C) is given by:
\[ \Lambda = \frac{\kappa}{C} \]
Rearranging gives:
\[ \kappa = \Lambda \times C = 200 \, \Omega^{-1} \, cm^2 \, mol^{-1} \times 0.02 \, mol/L = 4 \, \times \, 10^{-3} \, \Omega^{-1} \, cm^{-1} \]

### Step 2: Calculate the Cell Constant
Using the formula:
\[ \kappa = \frac{1}{R} \times G^* \]
Where \( G^* \) is the cell constant. Rearranging gives:
\[ G^* = \kappa \times R \]
Substituting the values for KCl:
\[ G^* = 4 \times 10^{-3} \, \Omega^{-1} \, cm^{-1} \times 100 \, \Omega = 0.4 \, cm^{-1} \]

### Step 3: Calculate the Specific Conductivity of Water
Given:
- Resistance of water = 10,000 \( \Omega \)

Using the formula:
\[ \kappa_{H_2O} = \frac{1}{R} \times G^* \]
Substituting the values:
\[ \kappa_{H_2O} = \frac{1}{10,000} \times 0.4 = 4 \times 10^{-5} \, \Omega^{-1} \, cm^{-1} \]

### Step 4: Calculate the Specific Conductivity of NaCl Solution
Given:
- Resistance of NaCl solution = 8,000 \( \Omega \)

Using the same formula:
\[ \kappa_{NaCl} = \frac{1}{R} \times G^* \]
Substituting the values:
\[ \kappa_{NaCl} = \frac{1}{8000} \times 0.4 = 5 \times 10^{-5} \, \Omega^{-1} \, cm^{-1} \]

### Step 5: Calculate the Conductivity of NaCl
The conductivity of the NaCl solution can be expressed as:
\[ \kappa_{NaCl \, solution} = \kappa_{NaCl} + \kappa_{H_2O} \]
Rearranging gives:
\[ \kappa_{NaCl} = \kappa_{NaCl \, solution} - \kappa_{H_2O} \]
Substituting the values:
\[ \kappa_{NaCl} = 5 \times 10^{-5} - 4 \times 10^{-5} = 1 \times 10^{-5} \, \Omega^{-1} \, cm^{-1} \]

### Step 6: Calculate the Molar Conductivity of NaCl
Using the formula:
\[ \Lambda_{NaCl} = \frac{\kappa_{NaCl}}{C} \]
Where \( C \) is the concentration in mol/L. We know:
- Moles of NaCl = \( \frac{585 \, g}{58.5 \, g/mol} = 10 \, mol \)

Thus, the concentration \( C \) can be expressed as:
\[ C = \frac{10 \, mol}{V} \]
Where \( V \) is the volume in liters.

Substituting this into the molar conductivity equation gives:
\[ 125 = \frac{1 \times 10^{-5} \times 1000}{\frac{10}{V}} \]
Rearranging gives:
\[ 125 = \frac{1 \times 10^{-2} \times V}{10} \]
\[ V = \frac{125 \times 10}{1 \times 10^{-2}} = 125 \times 10^2 = 12500 \, L \]

### Final Answer
The volume of water in the pool is \( 12500 \, L \).