The freezing point of a `5g CH_(3)COOH (aq) ` per `100g` water is `01.576^(@)C`. The van't Hoff factor `(K_(f)` of water `-1.86K mol^(-1)kg) :`

Options

A. `0.996`
B. 2
C. `0.5`
D. `1.016`

Explanation

To solve the problem, we need to find the van't Hoff factor (i) for acetic acid (CH₃COOH) using the freezing point depression formula. Here’s a step-by-step breakdown:

### Step 1: Understand the Freezing Point Depression Formula
The freezing point depression can be calculated using the formula:
\[ \Delta T_f = K_f \times m \times i \]
Where:
- \(\Delta T_f\) = depression in freezing point
- \(K_f\) = freezing point depression constant
- \(m\) = molality of the solution
- \(i\) = van't Hoff factor

### Step 2: Calculate the Depression in Freezing Point
Given that the freezing point of pure water is 0°C and the freezing point of the solution is 1.576°C, we can calculate \(\Delta T_f\):
\[ \Delta T_f = 0 - 1.576 = -1.576°C \]
(Note: The negative sign indicates a depression in freezing point.)

### Step 3: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the number of moles of acetic acid (CH₃COOH).

1. **Given mass of CH₃COOH**: 5 g
2. **Molecular mass of CH₃COOH**: 60 g/mol

Calculate the number of moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \, \text{g}}{60 \, \text{g/mol}} = \frac{1}{12} \, \text{mol} \]

Now, convert the mass of water from grams to kilograms:
\[ \text{Mass of water} = 100 \, \text{g} = 0.1 \, \text{kg} \]

Now, calculate the molality:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{1/12 \, \text{mol}}{0.1 \, \text{kg}} = \frac{1}{12 \times 0.1} = \frac{1}{1.2} \approx 0.833 \, \text{mol/kg} \]

### Step 4: Substitute Values into the Freezing Point Depression Formula
Now we can substitute the known values into the freezing point depression formula:
\[ -1.576 = 1.86 \times 0.833 \times i \]

### Step 5: Solve for the van't Hoff Factor (i)
Rearranging the formula to solve for \(i\):
\[ i = \frac{-1.576}{1.86 \times 0.833} \]

Calculating the denominator:
\[ 1.86 \times 0.833 \approx 1.549 \]

Now substituting back:
\[ i = \frac{-1.576}{1.549} \approx 1.016 \]

### Conclusion
The van't Hoff factor (i) for acetic acid in this solution is approximately 1.016.