3. If 122.3 grams of strontium chloride is mixed with 925.0 g of water, boiling point of the resulting solution be?

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**Question 3:**

If 122.3 grams of strontium chloride is mixed with 925.0 grams of water, what will the freezing and boiling point of the resulting solution be?

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**Description:**

This problem involves finding the freezing and boiling points of a solution when a given amount of strontium chloride (SrCl₂) is dissolved in a given mass of water (H₂O). This is a typical colligative properties problem, which depends on the number of particles in the solution rather than the type of particles.

**Key Concepts:**

1. **Freezing Point Depression:** When a solute is dissolved in a solvent, the freezing point of the solvent decreases. This can be calculated using the formula:
   
   \[
   \Delta T_f = i \cdot K_f \cdot m
   \]
   
   Where:
   - \(\Delta T_f\) is the freezing point depression.
   - \(i\) is the van't Hoff factor (number of particles the solute splits into).
   - \(K_f\) is the freezing point depression constant of the solvent.
   - \(m\) is the molality of the solution.
   
2. **Boiling Point Elevation:** When a solute is dissolved in a solvent, the boiling point of the solvent increases. This can be calculated using the formula:
   
   \[
   \Delta T_b = i \cdot K_b \cdot m
   \]
   
   Where:
   - \(\Delta T_b\) is the boiling point elevation.
   - \(i\) is the van't Hoff factor.
   - \(K_b\) is the boiling point elevation constant of the solvent.
   - \(m\) is the molality of the solution.

**Solution Steps:**

1. **Calculate molality (\( m \)) of the solution.**

2. **Determine the van't Hoff factor (\( i \)) for strontium chloride (SrCl₂).**

3. **Use the provided constants \( K_f \) and \( K_b \) for water.**

4. **Apply the formulas for freezing point depression and boiling point elevation to find the new freezing and boiling points.**

---

This example demonstrates how to use the properties of solutions to determine changes in physical properties like boiling and freezing points when a solute is added to a solvent.
Transcribed Image Text:**Question 3:** If 122.3 grams of strontium chloride is mixed with 925.0 grams of water, what will the freezing and boiling point of the resulting solution be? --- **Description:** This problem involves finding the freezing and boiling points of a solution when a given amount of strontium chloride (SrCl₂) is dissolved in a given mass of water (H₂O). This is a typical colligative properties problem, which depends on the number of particles in the solution rather than the type of particles. **Key Concepts:** 1. **Freezing Point Depression:** When a solute is dissolved in a solvent, the freezing point of the solvent decreases. This can be calculated using the formula: \[ \Delta T_f = i \cdot K_f \cdot m \] Where: - \(\Delta T_f\) is the freezing point depression. - \(i\) is the van't Hoff factor (number of particles the solute splits into). - \(K_f\) is the freezing point depression constant of the solvent. - \(m\) is the molality of the solution. 2. **Boiling Point Elevation:** When a solute is dissolved in a solvent, the boiling point of the solvent increases. This can be calculated using the formula: \[ \Delta T_b = i \cdot K_b \cdot m \] Where: - \(\Delta T_b\) is the boiling point elevation. - \(i\) is the van't Hoff factor. - \(K_b\) is the boiling point elevation constant of the solvent. - \(m\) is the molality of the solution. **Solution Steps:** 1. **Calculate molality (\( m \)) of the solution.** 2. **Determine the van't Hoff factor (\( i \)) for strontium chloride (SrCl₂).** 3. **Use the provided constants \( K_f \) and \( K_b \) for water.** 4. **Apply the formulas for freezing point depression and boiling point elevation to find the new freezing and boiling points.** --- This example demonstrates how to use the properties of solutions to determine changes in physical properties like boiling and freezing points when a solute is added to a solvent.
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