A 0.10 M solution of a monoprotic acid (`d=1.01g//cm^(3)`) is 5% dissociated what is the freezing point of the solution the molar mass of the acid is 300 and `K_(f)(H_(2)O)=1.86C//m`

Options

A. `-0.189^(@)C`
B. `-0.194^(@)C`
C. `-0.199^(@)C`
D. none of these

Answer

To solve the problem, we need to find the freezing point of a 0.10 M solution of a monoprotic acid that is 5% dissociated. We will use the formula for freezing point depression and the relationship between molality and molarity.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Molarity (M) = 0.10 M
- Density (d) = 1.01 g/cm³
- Molar mass of the acid (M_m) = 300 g/mol
- Degree of dissociation (α) = 5% = 0.05
- Freezing point depression constant (K_f) = 1.86 °C/m

2. **Calculate the Molality (m):**
- The formula to convert molarity to molality is:
\[ m = \frac{M \times 1000}{d \times (1000 - M_m)} \]
- Substitute the values:
\[ m = \frac{0.10 \times 1000}{1.01 \times (1000 - 300)} \]
- Calculate the denominator:
\[ 1000 - 300 = 700 \]
- Now calculate:
\[ m = \frac{100}{1.01 \times 700} = \frac{100}{707} \approx 0.141 mol/kg \]

3. **Calculate the Van 't Hoff Factor (i):**
- The van 't Hoff factor (i) for a monoprotic acid is given by:
\[ i = \alpha + 1 \]
- Substitute the value of α:
\[ i = 0.05 + 1 = 1.05 \]

4. **Calculate the Freezing Point Depression (ΔTf):**
- The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Substitute the values:
\[ \Delta T_f = 1.05 \cdot 1.86 \cdot 0.141 \]
- Calculate:
\[ \Delta T_f \approx 1.05 \cdot 1.86 \cdot 0.141 \approx 0.199 °C \]

5. **Calculate the Freezing Point of the Solution (Tf):**
- The freezing point of the solution is given by:
\[ T_f = 0 - \Delta T_f \]
- Substitute the value of ΔTf:
\[ T_f = 0 - 0.199 = -0.199 °C \]

### Final Answer:
The freezing point of the solution is approximately **-0.199 °C**.