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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