`A 0.01` molal solution of ammonia freezes at `-0.02^(@)C`. Calculate the van't Hoff factor, `i` and the percentage dissociation of ammonia in water. `(K_(f(H_(2O))))=1.86` deg `"molal"^(-1)`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Molality of ammonia solution (m) = 0.01 molal
- Freezing point depression (ΔTf) = -0.02 °C (the absolute value is 0.02 °C)
- Cryoscopic constant of water (Kf) = 1.86 °C kg/mol

### Step 2: Use the formula for freezing point depression
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) = freezing point depression
- \( i \) = van't Hoff factor
- \( K_f \) = cryoscopic constant
- \( m \) = molality of the solution

### Step 3: Rearrange the formula to solve for \( i \)
Rearranging the formula gives:
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ i = \frac{0.02}{1.86 \cdot 0.01} \]

### Step 5: Calculate \( i \)
Calculating the denominator:
\[ 1.86 \cdot 0.01 = 0.0186 \]
Now substituting back:
\[ i = \frac{0.02}{0.0186} \approx 1.075 \]

### Step 6: Calculate the percentage dissociation of ammonia
To find the percentage dissociation, we use the formula:
\[ \alpha = \frac{i - 1}{n - 1} \]
Where:
- \( \alpha \) = degree of dissociation
- \( n \) = number of particles the solute dissociates into (for ammonia, it dissociates into 2 particles: \( NH_4^+ \) and \( OH^- \), so \( n = 2 \))

### Step 7: Substitute the values into the dissociation formula
Substituting the values we have:
\[ \alpha = \frac{1.075 - 1}{2 - 1} = 0.075 \]

### Step 8: Calculate the percentage dissociation
To find the percentage dissociation:
\[ \text{Percentage dissociation} = \alpha \cdot 100 = 0.075 \cdot 100 = 7.5\% \]

### Final Results
- Van't Hoff factor \( i \approx 1.075 \)
- Percentage dissociation of ammonia in water = 7.5%