Math Problem Statement

Four liquids are described in the table below. Use the second column of the table to explain the order of their freezing points, and the third column to explain the order of their boiling points. For example, select '1' in the second column next to the liquid with the lowest freezing point. Select '2' in the second column next to the liquid with the next higher freezing point, and so on. In the third column, select '1' next to the liquid with the lowest boiling point, '2' next to the liquid with the next higher boiling point, and so on. Note: the density of water is 1.00 g/ml . solution freezing point boiling point 3.1 g of glycerin (C3H8O3) dissolved in 350. mL of water 3.1 g of sodium bromide (NaBr) dissolved in 350. mL of water 3.1 g of hydrochloric acid (HCl) dissolved in 350. mL of water 350. mL of pure water

Solution

To determine the order of freezing points and boiling points, we must consider colligative properties, which depend on the number of solute particles in solution, not their identity. These properties include freezing point depression and boiling point elevation.


Step-by-Step Analysis:

1. Freezing Point Order (Lowest to Highest)

  • Freezing Point Depression Formula: ΔTf=iKfm\Delta T_f = i \cdot K_f \cdot m Where:
    • ii = van 't Hoff factor (number of particles the solute dissociates into)

    • KfK_f = freezing point depression constant for water

    • mm = molality of the solution

    • Pure Water: Freezing point = 0C0^\circ \text{C} (highest).

    • Glycerin (i=1i = 1): It does not dissociate, so freezing point depression is small.

    • NaBr (i=2i = 2): Dissociates into 2 particles (Na+\text{Na}^+ and Br\text{Br}^-), causing a larger depression.

    • HCl (i=2i = 2): Dissociates into H+\text{H}^+ and Cl\text{Cl}^-, similar to NaBr, but slightly stronger due to full dissociation.

Order for Freezing Points (Lowest to Highest):

  1. HCl
  2. NaBr
  3. Glycerin
  4. Pure Water

2. Boiling Point Order (Lowest to Highest)

  • Boiling Point Elevation Formula: ΔTb=iKbm\Delta T_b = i \cdot K_b \cdot m Where:
    • ii, KbK_b, and mm have the same roles as in freezing point depression.

    • Pure Water: Boiling point = 100C100^\circ \text{C} (lowest).

    • Glycerin (i=1i = 1): Slight boiling point elevation due to no dissociation.

    • NaBr (i=2i = 2): Greater boiling point elevation.

    • HCl (i=2i = 2): Similar to NaBr, slightly stronger due to full dissociation.

Order for Boiling Points (Lowest to Highest):

  1. Pure Water
  2. Glycerin
  3. NaBr
  4. HCl

Final Table:

SolutionFreezing Point (Order)Boiling Point (Order)
3.1 g of glycerin (C3H8O3) in 350 mL water32
3.1 g of sodium bromide (NaBr) in 350 mL water23
3.1 g of hydrochloric acid (HCl) in 350 mL water14
350 mL of pure water41

Would you like further clarification on any of the steps or concepts?
Here are five additional questions to consider:

  1. What is the van 't Hoff factor for other common solutes?
  2. How do colligative properties differ for ionic vs. non-ionic solutes?
  3. How would increasing the amount of solute affect freezing/boiling points?
  4. Why does water have high colligative property constants (KfK_f, KbK_b)?
  5. What other colligative properties are influenced by solutes?

Tip: Always check if a solute fully dissociates when calculating colligative properties!

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Math Problem Analysis

Mathematical Concepts

Colligative Properties
Freezing Point Depression
Boiling Point Elevation

Formulas

Freezing Point Depression Formula: ΔTf = i × Kf × m
Boiling Point Elevation Formula: ΔTb = i × Kb × m

Theorems

Colligative property principles

Suitable Grade Level

Grades 10-12