Select the one with the highest boiling point. Select the one with the lowest freezing point. A. 1.0 m KNO3 B. 0.75 m NaCl C. 0.75 m CuCl₂ D. 2.0 m C12H22O11 (sucrose) E. pure water Determine the freezing point of a solution that contains 0.31 mol of sucrose in 175 g of water, where K₁ = 1.86°C/m. A. 3.3°C Your answer: B. 1.1°C C. 0.0°C D. -1.1°C E.-3.3°C

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### Concepts of Freezing Point Depression and Boiling Point Elevation

**Instructions:**

**1. Select the one with the highest boiling point.**
**2. Select the one with the lowest freezing point.**

Options: 
- A. 1.0 m KNO₃
- B. 0.75 m NaCl
- C. 0.75 m CuCl₂
- D. 2.0 m C₁₂H₂₂O₁₁ (sucrose)
- E. Pure water

**Problem:**
Determine the freezing point of a solution that contains 0.31 mol of sucrose in 175 g of water, where \( K_f = 1.86^{\circ}C/m \).

Your answer choices:
- A. 3.3°C
- B. 1.1°C
- C. 0.0°C
- D. –1.1°C
- E. –3.3°C

**Solution:**

To find the freezing point of the solution, use the formula for freezing point depression:

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

Where:
- \( \Delta T_f \) is the freezing point depression.
- \( i \) is the van't Hoff factor (which is 1 for sucrose since it does not ionize in solution).
- \( K_f \) is the cryoscopic constant of the solvent.
- \( m \) is the molality of the solution.

First, calculate the molality (\( m \)) of the solution:
\[ m = \frac{moles \, of \, solute}{kg \, of \, solvent} \]

Moles of sucrose = 0.31 mol
Mass of water = 175 g = 0.175 kg

\[ m = \frac{0.31 \, mol}{0.175 \, kg} \approx 1.77 \, m \]

Next, calculate the freezing point depression:
\[ \Delta T_f = 1 \times 1.86^{\circ}C/m \times 1.77 \, m \]

\[ \Delta T_f \approx 3.3^{\circ}C \]

Since this is the depression of the freezing point, subtract it from the normal freezing point of pure water (0°C):
\[ Final \,
Transcribed Image Text:### Concepts of Freezing Point Depression and Boiling Point Elevation **Instructions:** **1. Select the one with the highest boiling point.** **2. Select the one with the lowest freezing point.** Options: - A. 1.0 m KNO₃ - B. 0.75 m NaCl - C. 0.75 m CuCl₂ - D. 2.0 m C₁₂H₂₂O₁₁ (sucrose) - E. Pure water **Problem:** Determine the freezing point of a solution that contains 0.31 mol of sucrose in 175 g of water, where \( K_f = 1.86^{\circ}C/m \). Your answer choices: - A. 3.3°C - B. 1.1°C - C. 0.0°C - D. –1.1°C - E. –3.3°C **Solution:** To find the freezing point of the solution, use the formula for freezing point depression: \[ \Delta T_f = i \times K_f \times m \] Where: - \( \Delta T_f \) is the freezing point depression. - \( i \) is the van't Hoff factor (which is 1 for sucrose since it does not ionize in solution). - \( K_f \) is the cryoscopic constant of the solvent. - \( m \) is the molality of the solution. First, calculate the molality (\( m \)) of the solution: \[ m = \frac{moles \, of \, solute}{kg \, of \, solvent} \] Moles of sucrose = 0.31 mol Mass of water = 175 g = 0.175 kg \[ m = \frac{0.31 \, mol}{0.175 \, kg} \approx 1.77 \, m \] Next, calculate the freezing point depression: \[ \Delta T_f = 1 \times 1.86^{\circ}C/m \times 1.77 \, m \] \[ \Delta T_f \approx 3.3^{\circ}C \] Since this is the depression of the freezing point, subtract it from the normal freezing point of pure water (0°C): \[ Final \,
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