State True/False :
Van't Hoff factor for `10^(-3) M CH_(3)COOH` is `39//35 (wedge^(@)._(m)=350 S cm^(2) mol^(-1)` and `k=4xx10^(-5) S cm^(-1)).`

Explanation

To determine whether the statement regarding the Van't Hoff factor for \(10^{-3} M \, CH_3COOH\) is true or false, we will follow these steps:

### Step 1: Understand the Dissociation of Acetic Acid
Acetic acid (\(CH_3COOH\)) dissociates in water as follows:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
This means that for every mole of acetic acid that dissociates, one mole of acetate ion and one mole of hydrogen ion are produced.

### Step 2: Define the Van't Hoff Factor
The Van't Hoff factor (\(i\)) is defined as:
\[ i = 1 + \alpha \]
where \(\alpha\) is the degree of dissociation of the solute.

### Step 3: Calculate the Degree of Dissociation (\(\alpha\))
The degree of dissociation can be calculated using the formula:
\[ \alpha = \frac{\text{Molar conductivity}}{\text{Molar conductivity at infinite dilution}} \]
Given:
- Molar conductivity (\(\kappa\)) = \(4 \times 10^{-5} \, S \, cm^{-1}\)
- Concentration (\(C\)) = \(10^{-3} \, mol \, L^{-1}\)

First, we calculate the molar conductivity:
\[ \text{Molar conductivity} = \frac{\kappa \times 1000}{C} = \frac{4 \times 10^{-5} \times 1000}{10^{-3}} = 40 \, S \, cm^2 \, mol^{-1} \]

### Step 4: Use the Given Infinite Dilution Value
The infinite dilution value is given as \(350 \, S \, cm^2 \, mol^{-1}\).

Now, we can calculate \(\alpha\):
\[ \alpha = \frac{40}{350} = \frac{4}{35} \]

### Step 5: Calculate the Van't Hoff Factor (\(i\))
Now, substituting \(\alpha\) back into the equation for \(i\):
\[ i = 1 + \alpha = 1 + \frac{4}{35} = \frac{35}{35} + \frac{4}{35} = \frac{39}{35} \]

### Conclusion
The calculated Van't Hoff factor is \(\frac{39}{35}\), which matches the statement given in the question. Therefore, the statement is **True**.

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