A solution containing `8.0 g` of nicotine in `92 g` of water-frezes `0.925` degrees below the normal freezing point of water. If the freezing point depression constant `K_(f) = -1.85^(@)C mol^(-1)` then the molar mass of nicotine is -

Options

A. 16
B. 80
C. 320
D. 160

Answer

To find the molar mass of nicotine based on the given data, we can follow these steps:

### Step 1: Identify the given data
- Mass of nicotine (solute), \( W_b = 8.0 \, \text{g} \)
- Mass of water (solvent), \( W_a = 92.0 \, \text{g} \)
- Freezing point depression, \( \Delta T_f = 0.925 \, \text{°C} \)
- Freezing point depression constant for water, \( K_f = 1.85 \, \text{°C kg/mol} \)

### Step 2: Convert the mass of the solvent to kilograms
Since the freezing point depression constant \( K_f \) is given in °C kg/mol, we need to convert the mass of water from grams to kilograms:
\[ W_a = 92.0 \, \text{g} = \frac{92.0}{1000} \, \text{kg} = 0.092 \, \text{kg} \]

### Step 3: Use the formula for freezing point depression
The formula relating freezing point depression to molality is:
\[ \Delta T_f = K_f \times m \]
where \( m \) is the molality of the solution.

### Step 4: Calculate molality
Molality \( m \) is defined as:
\[ m = \frac{n}{W_a} \]
where \( n \) is the number of moles of solute. The number of moles \( n \) can be expressed in terms of the molar mass \( M \):
\[ n = \frac{W_b}{M} \]
Thus, we can write:
\[ m = \frac{W_b}{M \times W_a} \]

### Step 5: Substitute into the freezing point depression formula
Substituting the expression for molality into the freezing point depression formula gives:
\[ \Delta T_f = K_f \times \frac{W_b}{M \times W_a} \]

### Step 6: Rearranging to find the molar mass
Rearranging the equation to solve for \( M \):
\[ M = \frac{K_f \times W_b}{\Delta T_f \times W_a} \]

### Step 7: Plug in the values
Now, substituting the known values into the equation:
\[ M = \frac{1.85 \, \text{°C kg/mol} \times 8.0 \, \text{g}}{0.925 \, \text{°C} \times 0.092 \, \text{kg}} \]

### Step 8: Calculate the molar mass
Calculating the numerator:
\[ 1.85 \times 8.0 = 14.8 \, \text{°C kg} \]
Calculating the denominator:
\[ 0.925 \times 0.092 = 0.08515 \, \text{°C kg} \]
Now, substituting these values:
\[ M = \frac{14.8}{0.08515} \approx 173.91 \, \text{g/mol} \]

### Conclusion
The molar mass of nicotine is approximately \( 173.91 \, \text{g/mol} \).

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