A 0.10 M solution of a mono protic 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 determine 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, which is given by:

\[ \Delta T_f = K_f \cdot m \cdot i \]

where:
- \(\Delta T_f\) is the depression in freezing point,
- \(K_f\) is the freezing point depression constant,
- \(m\) is the molality of the solution,
- \(i\) is the Van't Hoff factor.

### Step 1: Calculate the mass of the solution
Given:
- Density of the solution, \(d = 1.01 \, \text{g/cm}^3\)
- Volume of the solution = 1 L = 1000 cm³

Using the formula for mass:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.01 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1010 \, \text{g} \]

### Step 2: Calculate the mass of the solute
Given:
- Molarity of the solution = 0.10 M
- Molar mass of the acid = 300 g/mol

Using the formula for moles:
\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume} = 0.10 \, \text{mol/L} \times 1 \, \text{L} = 0.10 \, \text{mol} \]

Now, calculate the mass of the solute:
\[ \text{Mass of solute} = \text{Moles} \times \text{Molar mass} = 0.10 \, \text{mol} \times 300 \, \text{g/mol} = 30 \, \text{g} \]

### Step 3: Calculate the mass of the solvent
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1010 \, \text{g} - 30 \, \text{g} = 980 \, \text{g} \]

### Step 4: Calculate the molality of the solution
Molality (\(m\)) is defined as moles of solute per kilogram of solvent:
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.10 \, \text{mol}}{0.980 \, \text{kg}} \approx 0.102 \, \text{mol/kg} \]

### Step 5: Calculate the Van't Hoff factor (\(i\))
Given that the acid is 5% dissociated:
\[ \alpha = 0.05 \quad (\text{5% dissociation}) \]
\[ i = 1 + \alpha = 1 + 0.05 = 1.05 \]

### Step 6: Calculate the depression in freezing point (\(\Delta T_f\))
Using the freezing point depression formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Substituting the values:
\[ \Delta T_f = 1.86 \, \text{°C/m} \cdot 0.102 \, \text{mol/kg} \cdot 1.05 \]
\[ \Delta T_f \approx 0.199 \, \text{°C} \]

### Step 7: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C, so:
\[ \text{Freezing point of solution} = 0 \, \text{°C} - \Delta T_f = 0 \, \text{°C} - 0.199 \, \text{°C} = -0.199 \, \text{°C} \]

### Final Answer
The freezing point of the solution is approximately \(-0.199 \, \text{°C}\).

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