CHEM 120 - jan 15

Objective: Calculate the difference in freezing point caused by a solute in a solvent, specifically glucose in water.
- Formula Used: $ext{Delta} T<em>f = K</em>f imes ext{molality}$
- Where:
  - $K_f$ is the cryoscopic constant for the solvent.
  - $ext{molality} = rac{ ext{moles of solute}}{ ext{kilograms of solvent}}$
Known Variables:
- $K_f$ for water = 1.86 degrees per molality.
- Moles of glucose: 0.5000 moles.

Given: Grams of glucose needed to calculate moles, not provided directly.
Conversion: Use molar mass of glucose (C6H12O6) for this calculation:
- $ext{Molar mass of glucose} = 180.18 ext{ g/mol}$
Given 1.6 liters of solution, converted to milliliters: 1.6 L = 1600 mL.

Determine total mass of the solution:
- Using the density, mass of solution = density (g/mL) x volume (mL)
- Density assumed for water = 1 g/mL thus:
- $1600 ext{ mL} imes 1 ext{ g/mL} = 1600 ext{ g of solution}$
Mass of solute (glucose) = 90.08 grams, so:
- Mass of solvent (water) = total mass - mass of solute
- $1600 ext{ g} - 90.08 ext{ g} = 1509.92 ext{ g}$
Convert to kilograms:
- $rac{1509.92 ext{ g}}{1000} = 1.50992 ext{ kg}$

Using values gathered:
- $ext{molality} = rac{ ext{moles of glucose}}{ ext{kilograms of water}} = rac{0.5000 ext{ moles}}{1.57392 ext{ kg}} = 0.317679$

Apply values to the freezing point depression formula:
- $ext{Delta} T_f = 1.86 imes 0.317679 = 0.59 ext{ degrees}$
Resulting freezing point of the solution:
- Normal freezing point of water is 0 degrees Celsius:
- $0 - 0.59 = -0.59 ext{ degrees Celsius}$

Another given scenario: Dissolving 10 g of quinine in 50 mL of ethanol results in a freezing point depression of 1.55 degrees C.
Known: $K_f$ for ethanol = 1.99 degrees per molality.

Rearranging the freezing point depression formula:
- $1.55 = 1.99 imes molality$
Solving for molality:
- $molality = rac{1.55}{1.99} = 0.779 ext{ mol/kg}$

Density for ethanol = 0.789 g/mL. Thus for 50 mL:
- $ext{mass} = 0.789 ext{ g/mL} imes 50 ext{ mL} = 39.45 ext{ g}$
Convert to kilograms:
- $rac{39.45}{1000} = 0.03945 ext{ kg}$

Find moles of quinine:
- Using molality formula:
- $0.779 = rac{ ext{moles}}{0.03945 ext{ kg}}$ \
- $ext{moles} = 0.779 imes 0.03945 = 0.0308 ext{ moles}$

Molar mass = grams of solute/moles of solute:
- ext{molar mass} = rac{10 ext{ g}}{0.0308 ext{ moles}} \
  = 325.32 ext{ g/mol}

Concept Introduction: Boiling point elevation is the phenomenon where the boiling point of a solution is higher than that of the pure solvent due to the presence of solute.
- Formula: $ext{Delta} T<em>b = K</em>b imes m$ where:
- $K_b$ is the boiling point elevation constant.
- $m$ is the molality of the solution.

Example calculation for ethylene glycol, $K_b = 0.51 ext{ degrees per molality}$
- Molality established previously: 5.32
- $ext{Delta} T_b = 0.51 imes 5.32 = 2.72 ext{ degrees}$
- New boiling point = 100 degrees + 2.7 degrees = 102.7 degrees.

Colligative properties depend only on the number of solute particles and not their identity.
Key Concept: A solution containing more solute particles will exhibit a lower freezing point compared to a solution with fewer solute particles.
Van 't Hoff Factor (i): Accounts for dissociation of electrolytes in solutions:
- $i = rac{ ext{actual number of particles}}{ ext{initial units dissolved}}$
Examples:
- Sodium Chloride (NaCl): dissociates into Na+ and Cl-, thus i = 2.
- Sodium Sulfate (Na2SO4): dissociates into Na+ (2) and SO4^(2-), thus i = 3.

Freezing and boiling point calculations reveal the effects of solutes on solvents.
The analysis focuses on how changes in solute concentration impact the physical properties of the resulting solutions, emphasizing the underlying principles of colligative properties and their quantitative outcomes in both freezing point depression and boiling point elevation.