Assuming 100% polymerization of an organic compound in its aqueous solution, the number of moles of organic compound undergoing polymerization containing `9.4g` of organic compound per `100g` of the solvent is ……….Freezing point depression is `0.93K, K_(f)` of water is `1.86K kg"mol"^(-1)` and molecular weight of organic compound is `94u`

Explanation

To solve the problem, we need to find the number of moles of the organic compound undergoing polymerization. We will use the freezing point depression formula and the given data to find the answer step by step.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of organic compound (solute) = 9.4 g
- Mass of solvent (water) = 100 g
- Freezing point depression (ΔTf) = 0.93 K
- Freezing point depression constant (Kf) of water = 1.86 K kg/mol
- Molecular weight of the organic compound = 94 g/mol

2. **Convert the Mass of Solvent to kg:**
Since Kf is given in K kg/mol, we need to convert the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent} = 100 \text{ g} = 0.1 \text{ kg} \]

3. **Use the Freezing Point Depression Formula:**
The formula for freezing point depression is:
\[ \Delta Tf = i \cdot Kf \cdot m \]
where:
- \( \Delta Tf \) = freezing point depression
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( Kf \) = freezing point depression constant
- \( m \) = molality of the solution

4. **Calculate the Molality (m):**
Molality is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the number of moles of the organic compound:
\[ \text{Number of moles of organic compound} = \frac{\text{mass}}{\text{molecular weight}} = \frac{9.4 \text{ g}}{94 \text{ g/mol}} = 0.1 \text{ moles} \]
Now, calculate the molality:
\[ m = \frac{0.1 \text{ moles}}{0.1 \text{ kg}} = 1 \text{ mol/kg} \]

5. **Substitute Values into the Freezing Point Depression Formula:**
Rearranging the formula to find \( i \):
\[ i = \frac{\Delta Tf}{Kf \cdot m} \]
Substitute the known values:
\[ i = \frac{0.93 \text{ K}}{1.86 \text{ K kg/mol} \cdot 1 \text{ mol/kg}} = 0.5 \]

6. **Determine the Number of Moles of the Organic Compound Undergoing Polymerization:**
The van 't Hoff factor \( i \) is related to the number of particles formed from the solute. Since \( i = \frac{1}{n} \), we can find \( n \):
\[ n = \frac{1}{i} = \frac{1}{0.5} = 2 \]
This means that 2 moles of the organic compound are formed from 1 mole of the original compound due to polymerization.

### Final Answer:
The number of moles of organic compound undergoing polymerization is **2 moles**.