Video summary

Sifat Koligatif 2Penurunan Tekanan Uap

Main summary

Key takeaways

Educational

Main ideas & concepts (Colligative Properties: Decrease in Vapor Pressure)

  • Colligative properties of a solution depend only on the number of dissolved particles (not the type of solute).
    • Therefore, more solute particles dissolved → larger colligative effect.

Key comparison example (sugar solutions in water)

  • Two beakers contain the same amount of water (100 mL).
  • Beaker 1: 1 spoon sugar
  • Beaker 2: 2 spoons sugar
  • Beaker 2 has greater colligative properties because it has more dissolved particles.

The video focuses on the first colligative property:

  1. Decrease in vapor pressure (ΔP)
  2. Others mentioned (not developed in detail here):
    • increase in boiling point (ΔT_B)
    • decrease in freezing point (ΔT_F)
    • osmotic pressure (π or P)

Electrolyte vs. non-electrolyte (why particles differ)

  • Non-electrolyte: does not ionize, so it contributes one particle per formula unit.
  • Electrolyte (e.g., NaCl): ionizes, producing more particles than the original solute.

Because electrolytes create more particles, they show larger colligative effects.

van’t Hoff factor (i) concept

To account for ionization, the video introduces the van’t Hoff factor (i):

[ i = 1 + (n - 1)\cdot \alpha ]

  • n = number of ions produced per formula unit
  • α = degree of ionization

This factor modifies how to calculate vapor pressure decrease for electrolytes.

Decrease in vapor pressure (ΔP): what it means

Evaporation and solute effects

  • Evaporation is a phase change from liquid → gas (liquid molecules leave the surface).
  • During boiling, gas bubbles form due to evaporation.
  • If a non-volatile solute (e.g., sugar, salt, spices) is added:
    • solute particles block/prevent solvent molecules from escaping,
    • so boiling/evaporation is reduced,
    • meaning the vapor pressure becomes lower.

Vapor pressure and equilibrium

  • In a closed container, evaporation and condensation reach equilibrium.
  • The maximum stable vapor pressure at equilibrium is saturated vapor pressure.
  • For a pure solvent: saturated vapor pressure is P₀.
  • For a solution: saturated vapor pressure is P.
  • Since solute reduces evaporation:

[ P(\text{solution}) < P_0(\text{pure solvent}) ]

This phenomenon is called vapor pressure reduction.

Definition of decrease in vapor pressure

[ \Delta P = P_0 - P ]

Raoult’s law (Roll’s law) and vapor pressure formulas

For non-electrolyte solutions

[ P = X_{\text{solvent}}\cdot P_0 ]

Since:

[ X_{\text{solvent}} + X_{\text{solute}} = 1 \quad \Rightarrow \quad X_{\text{solvent}} = 1 - X_{\text{solute}} ]

Then:

[ \Delta P = P_0 - P ] [ \Delta P = P_0 - (X_{\text{solvent}}\cdot P_0) ] [ \Delta P = X_{\text{solute}}\cdot P_0 ]

So for non-electrolytes, the decrease can be calculated either as:

  • ΔP = P₀ − P, or
  • ΔP = X_solute · P₀

For electrolyte solutions

Because ionization increases particle number, the electrolyte case uses the van’t Hoff factor (i).

Conceptually (as presented in the video):

  • ΔP depends on P₀, the solute mole fraction, and i (structure may vary depending on how the ratio is written).

The key point: include i for electrolytes.

Worked examples & numerical steps

Example 1: Urea (non-electrolyte)

Given:

  • P₀ = 100 mmHg
  • Urea is a non-electrolyte, so no van’t Hoff factor is needed (i = 1 implicitly)
  • “Urea mole fraction 10%” → X_solute = 0.1

Compute: [ \Delta P = X_{\text{solute}}\cdot P_0 = 0.1 \cdot 100\,\text{mmHg} = 10\,\text{mmHg} ]

Result: vapor pressure decreases by 10 mmHg.

Example 2: Glucose (non-electrolyte) vapor pressure at 300°C

Given:

  • 2 moles glucose dissolved in 50 moles water
  • P₀ at 300°C = 31.80 mmHg (pure solvent)
  • Glucose is treated as a non-electrolyte

Steps:

  1. Mole fraction of solute: [ X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} ] [ X_{\text{solute}} = \frac{2}{2+50} = \frac{2}{52} = 0.038 ]

  2. Mole fraction of solvent: [ X_{\text{solvent}} = 1 - X_{\text{solute}} = 1 - 0.038 = 0.962 ]

  3. Raoult’s law: [ P = X_{\text{solvent}}\cdot P_0 = 0.962 \cdot 31.80 = 30.59\,\text{mmHg} ]

Result: vapor pressure of the glucose solution = 30.59 mmHg.

Instructional component (how to solve these problems)

Before using vapor pressure equations:

  1. Check whether the solute is an electrolyte or non-electrolyte.
  2. Choose the appropriate formula:

  • If non-electrolyte:

    • [ P = X_{\text{solvent}}\cdot P_0 ]

    • [ \Delta P = X_{\text{solute}}\cdot P_0 ]

  • If electrolyte:

    • include the van’t Hoff factor (i) in the vapor pressure decrease (because ionization increases particle count)

For mole fractions:

  • [ X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} ]

  • [ X_{\text{solvent}} = 1 - X_{\text{solute}} ]

Practice & wrap-up

  • The teacher provides practice questions (3 total) and encourages viewers to try them.
  • The video ends with standard greetings.

Speakers / sources featured

  • Mrs. Pil / the Chemistry instructor (main speaker/teacher)
  • Raoult (referenced for Raoult’s law; “Roll’s law” appears in subtitles)
  • van’t Hoff (referenced for van’t Hoff factor, ( i = 1 + (n-1)\alpha ))

Original video