When `250 mg` of eugenol is added to `100 g` of camphor `(k_(f)=39.7" molality"^(-1))`, it lowered the freezing point by `0.62^(@)C` . The molar of eugenol is `:`

Options

A. `1.6xx10^(2) g//mol`
B. `1.6xx10^(4)g//mol`
C. `1.6xx10^(3)g//mol`
D. `200g//mol`

Explanation

To find the molar mass of eugenol based on the information provided, we can use the formula related to freezing point depression:

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

Where:
- \(\Delta T_f\) is the depression in freezing point,
- \(K_f\) is the cryoscopic constant,
- \(m\) is the molality of the solution.

### Step 1: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We can express it as:

\[ m = \frac{n}{W_{solvent}} \]

Where:
- \(n\) is the number of moles of solute (eugenol),
- \(W_{solvent}\) is the mass of the solvent (camphor) in kilograms.

Given that the mass of camphor is \(100 \, g\), we convert it to kilograms:

\[ W_{solvent} = 100 \, g = 0.1 \, kg \]

### Step 2: Rearranging the freezing point depression formula
From the freezing point depression formula, we can express molality as:

\[ m = \frac{\Delta T_f}{K_f} \]

Substituting the values:

\[ m = \frac{0.62 \, °C}{39.7 \, °C \cdot kg/mol} \]

### Step 3: Calculate the molality
Now we can calculate the molality:

\[ m = \frac{0.62}{39.7} \approx 0.0156 \, mol/kg \]

### Step 4: Calculate the number of moles of eugenol
We know that:

\[ m = \frac{n}{W_{solvent}} \implies n = m \cdot W_{solvent} \]

Substituting the values:

\[ n = 0.0156 \, mol/kg \cdot 0.1 \, kg = 0.00156 \, mol \]

### Step 5: Convert the mass of eugenol to grams
The mass of eugenol is given as \(250 \, mg\). We convert this to grams:

\[ W_{solute} = 250 \, mg = 0.250 \, g \]

### Step 6: Calculate the molar mass of eugenol
The molar mass (M) can be calculated using the formula:

\[ M = \frac{W_{solute}}{n} \]

Substituting the values:

\[ M = \frac{0.250 \, g}{0.00156 \, mol} \approx 160.26 \, g/mol \]

### Final Answer
The molar mass of eugenol is approximately \(160 \, g/mol\).

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