Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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### Freezing Point Depression Problem

**Problem Statement:**
0.25 mol of a nonvolatile, nonionic solute dissolved in 300 g of water (\(K_f = 1.86\) °C·kg/mol) will lower the freezing point by how many degrees?

**Options:**
- O 0.65 °C
- O 1.6 °C
- O 2.2 °C
- O 3.2 °C

**Explanation:**

To solve this problem, we use the formula for freezing point depression:

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

where:
- \(\Delta T_f\) is the change in freezing point,
- \(K_f\) is the freezing point depression constant (\(1.86\) °C·kg/mol for water),
- \(m\) is the molality of the solution.

First, we need to calculate the molality (\(m\)) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent:

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

Given:
- Moles of solute = 0.25 mol
- Mass of water = 300 g = 0.300 kg

Now, calculate the molality:

\[
m = \frac{0.25 \text{ mol}}{0.300 \text{ kg}} = 0.833 \text{ mol/kg}
\]

Now apply the freezing point depression formula:

\[
\Delta T_f = 1.86 \text{ °C·kg/mol} \times 0.833 \text{ mol/kg} \approx 1.55 \text{ °C}
\]

After rounding, the freezing point will be lowered by approximately \(1.6\) °C. 

Thus, the correct answer is:

- O 1.6 °C
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Transcribed Image Text:### Freezing Point Depression Problem **Problem Statement:** 0.25 mol of a nonvolatile, nonionic solute dissolved in 300 g of water (\(K_f = 1.86\) °C·kg/mol) will lower the freezing point by how many degrees? **Options:** - O 0.65 °C - O 1.6 °C - O 2.2 °C - O 3.2 °C **Explanation:** To solve this problem, we use the formula for freezing point depression: \[\Delta T_f = K_f \cdot m\] where: - \(\Delta T_f\) is the change in freezing point, - \(K_f\) is the freezing point depression constant (\(1.86\) °C·kg/mol for water), - \(m\) is the molality of the solution. First, we need to calculate the molality (\(m\)) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \] Given: - Moles of solute = 0.25 mol - Mass of water = 300 g = 0.300 kg Now, calculate the molality: \[ m = \frac{0.25 \text{ mol}}{0.300 \text{ kg}} = 0.833 \text{ mol/kg} \] Now apply the freezing point depression formula: \[ \Delta T_f = 1.86 \text{ °C·kg/mol} \times 0.833 \text{ mol/kg} \approx 1.55 \text{ °C} \] After rounding, the freezing point will be lowered by approximately \(1.6\) °C. Thus, the correct answer is: - O 1.6 °C
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