A `1.00` molal aqueous solution of trichloroacetic acid `(CCl_(3)COOH)` is heated to its boiling point. The solution has the boiling point of `100.18^(@)C`. Determine the van't Hoff factor for trichloroacetic acid.
(`K_(b)` for water `=0.512K kg "mol"^(-1)`)

Answer

To determine the van't Hoff factor (i) for trichloroacetic acid (CCl₃COOH) in a 1.00 molal aqueous solution, we can follow these steps:

### Step 1: Determine the change in boiling point (ΔT_b)
The boiling point of pure water is 100.00 °C. The boiling point of the solution is given as 100.18 °C.

\[ \Delta T_b = \text{Boiling point of solution} - \text{Boiling point of water} \]

\[ \Delta T_b = 100.18 °C - 100.00 °C = 0.18 °C \]

### Step 2: Use the formula for boiling point elevation
The formula for boiling point elevation is given by:

\[ \Delta T_b = K_b \cdot m \cdot i \]

Where:
- \( K_b \) is the ebullioscopic constant (0.512 K kg/mol for water),
- \( m \) is the molality of the solution (1.00 molal),
- \( i \) is the van't Hoff factor.

### Step 3: Rearranging the formula to find i
We can rearrange the formula to solve for the van't Hoff factor (i):

\[ i = \frac{\Delta T_b}{K_b \cdot m} \]

### Step 4: Substitute the known values
Now we can substitute the known values into the equation:

\[ i = \frac{0.18 °C}{0.512 \, \text{K kg/mol} \cdot 1.00 \, \text{molal}} \]

### Step 5: Calculate i
Calculating the value:

\[ i = \frac{0.18}{0.512} \approx 0.3516 \]

### Step 6: Interpret the result
The van't Hoff factor (i) indicates the number of particles the solute breaks into when dissolved. Since trichloroacetic acid is a weak acid, it partially dissociates in solution.

### Final Answer
The van't Hoff factor (i) for trichloroacetic acid in a 1.00 molal aqueous solution is approximately 0.3516.

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