0.25 mol of a nonvolatile, nonionic solute dissolved in 300 g of water (Kf = 1.86) will lower the freezing point how many degrees? 0.65 °C O 1.6 °C O 2.2 °C 3.2 °C
0.25 mol of a nonvolatile, nonionic solute dissolved in 300 g of water (Kf = 1.86) will lower the freezing point how many degrees? 0.65 °C O 1.6 °C O 2.2 °C 3.2 °C
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07e46d5b-e450-4cd0-bbfa-546b13ff1c45%2Fb8cb16c8-1a31-4bea-ba2e-ec3deea92aca%2Fuegpf18_processed.jpeg&w=3840&q=75)
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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