An aqueous solution of a weak monobasic acid containing `0.1g`in `21.7`g of water freezes at`272.813K`.If the value of `K_(f)`for water is `1.86K//m`,what is the molecular mass of the monobasic acid

Options

A. `50g//mol`
B. `46g//mol`
C. `55g//mol`
D. `60g//mol`

Explanation

To find the molecular mass of the weak monobasic acid from the given data, we can follow these steps:

### Step 1: Calculate the change in freezing point (ΔTf)
The freezing point of pure water (T0f) is 273.15 K. The freezing point of the solution (Tf) is given as 272.813 K.

\[ \Delta Tf = T0f - Tf = 273.15 \, \text{K} - 272.813 \, \text{K} = 0.337 \, \text{K} \]

### Step 2: Use the formula for freezing point depression
The formula for freezing point depression is given by:

\[ \Delta Tf = K_f \cdot m \]

where:
- \( K_f \) is the freezing point depression constant (1.86 K/m for water),
- \( m \) is the molality of the solution.

### Step 3: Calculate molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We can rearrange the equation to find molality:

\[ m = \frac{\Delta Tf}{K_f} = \frac{0.337 \, \text{K}}{1.86 \, \text{K/m}} \approx 0.181 \, \text{mol/kg} \]

### Step 4: Calculate the number of moles of solute
We know that molality (m) is also defined as:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

The mass of the solvent (water) is given as 21.7 g, which is 0.0217 kg. Rearranging gives:

\[ \text{moles of solute} = m \cdot \text{mass of solvent in kg} = 0.181 \, \text{mol/kg} \times 0.0217 \, \text{kg} \approx 0.00393 \, \text{mol} \]

### Step 5: Calculate the molecular mass of the solute
The molecular mass (M) of the solute can be calculated using the formula:

\[ M = \frac{\text{mass of solute (g)}}{\text{moles of solute (mol)}} \]

Substituting the values:

\[ M = \frac{0.1 \, \text{g}}{0.00393 \, \text{mol}} \approx 25.5 \, \text{g/mol} \]

### Final Answer
The molecular mass of the monobasic acid is approximately **25.5 g/mol**.

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