A big irrengular shaped vessel contained waer the sp conductance of which was `2.56 xx 10^(-5) mho cm^(-1) 500g` of `NaCI` was then added to the water and the specific conductance after the addition of `NaCI` was found to be `3.10 xx 10^(-5) mho cm^(-1)`. find the capacity of the vessel if it is fulfilled with water `(lambda_(oo)NaCI = 149.9)`

Explanation

To solve the problem step by step, we will follow the approach outlined in the video transcript.

### Step 1: Define the volume of the vessel
Let the volume of the vessel be \( V \) liters.

### Step 2: Calculate the equivalent mass of NaCl
The equivalent mass of NaCl can be calculated using its molar mass. The molar mass of NaCl is approximately 58.5 g/mol. Therefore, the equivalent mass is also 58.5 g (since NaCl is a 1:1 electrolyte).

### Step 3: Calculate the volume containing one equivalent of NaCl
The volume containing one equivalent of NaCl can be calculated using the formula:

\[ \text{Volume containing one equivalent} = \frac{V}{\frac{500 \, \text{g}}{58.5 \, \text{g/equiv}}} \]

This simplifies to:

\[ \text{Volume containing one equivalent} = \frac{V}{8.547} \]

### Step 4: Calculate the specific conductance of NaCl solution
The specific conductance of the NaCl solution can be calculated by subtracting the specific conductance of water from the specific conductance of the NaCl solution:

\[ \kappa = \kappa_{\text{NaCl solution}} - \kappa_{\text{water}} \]

Substituting the given values:

\[ \kappa = 3.10 \times 10^{-5} \, \text{mho/cm} - 2.56 \times 10^{-5} \, \text{mho/cm} \]

Calculating this gives:

\[ \kappa = 0.54 \times 10^{-5} \, \text{mho/cm} \]

### Step 5: Relate equivalent conductance to specific conductance
The equivalent conductance (\( \lambda_{\text{equivalent}} \)) is related to specific conductance and the volume containing one equivalent:

\[ \lambda_{\text{equivalent}} = \kappa \times \text{Volume containing one equivalent} \]

For very dilute solutions, this is equal to the molar conductivity at infinite dilution (\( \lambda_{\infty} \)). Therefore, we can write:

\[ 149.9 = 0.54 \times 10^{-5} \times \frac{V}{8.547} \]

### Step 6: Solve for the volume \( V \)
Rearranging the equation to solve for \( V \):

\[ V = \frac{149.9 \times 8.547}{0.54 \times 10^{-5}} \]

Calculating this gives:

\[ V \approx 237258.38 \, \text{liters} \]

### Final Answer
The capacity of the vessel is approximately **237258.38 liters**.

---