A 12.0 g sample of a nonelectrolyte is dissolved in 80 g of water. The solution freezes at -1.94 degrees Celsius. Assuming ideal solution behavior, calculate the substance's molar mass

To calculate the molar mass of the nonelectrolyte, we can use the depression in the freezing point equation, which is based on the concept of freezing point depression:

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

Where:
- \(\Delta T_f\) is the freezing point depression (the difference between the freezing point of pure solvent and the freezing point of the solution),
- \(i\) is the van’t Hoff factor (for nonelectrolytes, \(i = 1\) since they do not dissociate),
- \(K_f\) is the freezing point depression constant of the solvent (for water, \(K_f = 1.86 \, ^\circ \text{C} \cdot \text{kg/mol}\)),
- \(m\) is the molality of the solution, defined as the number of moles of solute per kilogram of solvent.

### Step-by-Step Calculation

1. **Find the freezing point depression \(\Delta T_f\):**

   \[
   \Delta T_f = T_f^\circ - T_f = 0^\circ \text{C} - (-1.94^\circ \text{C}) = 1.94^\circ \text{C}
   \]

2. **Use the freezing point depression equation to find molality:**

   Rearrange the equation for molality \(m\):

   \[
   m = \frac{\Delta T_f}{i \cdot K_f} = \frac{1.94}{1 \cdot 1.86} \approx 1.043 \, \text{mol/kg}
   \]

3. **Calculate the number of moles of solute:**

   Molality is defined as the number of moles of solute per kilogram of solvent. Given the mass of water (solvent) is 80 g, or 0.080 kg, the number of moles of solute can be calculated as:

   \[
   \text{moles of solute} = m \times \text{kg of solvent} = 1.043 \, \text{mol/kg} \times 0.080 \, \text{kg} = 0.08344 \, \text{mol}
   \]

4. **Calculate the molar mass of the solute:**

   Molar mass is the mass of the solute divided by the number of moles of solute:

   \[
   \text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{12.0 \, \text{g}}{0.08344 \, \text{mol}} \approx 143.8 \, \text{g/mol}
   \]

### Final Answer:
The molar mass of the nonelectrolyte is approximately **143.8 g/mol**.

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Do you want more details or have any questions? 

Here are 5 related questions you might find interesting:
1. What would happen if the substance were an electrolyte?
2. How does the molar mass of a substance affect its properties in a solution?
3. Can you calculate the boiling point elevation for the same solution?
4. How does the choice of solvent impact freezing point depression?
5. What assumptions are made in ideal solution behavior?

**Tip:** Always check the units when working with colligative properties to avoid errors!

A 12.0 g sample of a nonelectrolyte is dissolved in 80 g of water. The solution freezes at -1.94 degrees Celsius. Assuming ideal solution behavior, calculate the substance's molar mass.

Learn how to calculate the molar mass of a nonelectrolyte using freezing point depression. This example involves a solution with a 12.0 g sample of solute dissolved in 80 g of water, freezing at -1.94°C. Suitable for advanced high school chemistry students, this step-by-step guide covers key colligative properties and the use of molality in freezing point depression.

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