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Chemistry: Solutions - Lecture 5 Class 12

Chemistry: Solutions - Lecture 5 | CHEMCA
Class 12 / JEE / NEET Chapter 1

Chemistry - Solutions | Lecture 5

By Abhishek Sengar

Lecture Chapters

Click any topic to jump directly to that point in the video.

Theory & Quick Notes

1 Intro to Colligative Properties

Colligative properties are properties of solutions that depend entirely on the ratio of the number of solute particles to the number of solvent particles in a solution, and not on the nature or identity of the chemical species present.

The 4 Main Colligative Properties:

  1. Relative Lowering of Vapor Pressure (RLVP)
  2. Elevation in Boiling Point (ΔTb)
  3. Depression in Freezing Point (ΔTf)
  4. Osmotic Pressure (π)

2 Relative Lowering of Vapor Pressure (RLVP)

When a non-volatile solute is added to a volatile solvent, the vapor pressure of the solvent decreases. The relative lowering of vapor pressure is directly equal to the mole fraction of the solute.

Exact Formula

(P°A - PS) / P°A = xB

Where xB is the mole fraction of the solute: nB / (nA + nB)

DILUTE RULE

For Dilute Solutions

If nA >> nB, then (nA + nB) ≈ nA

(P°A - PS) / P°A ≈ nB / nA

3 Boiling Point & Freezing Point Changes

Elevation in Boiling Point (ΔTb)

Adding a non-volatile solute lowers the vapor pressure. Thus, a higher temperature is required for the vapor pressure to equal 1 atm (external pressure). Result: Boiling point increases.

ΔTb = Tb - T°b = Kb × m

Kb = Ebullioscopic Constant. m = molality.

Depression in Freezing Point (ΔTf)

The freezing point is the temperature where the vapor pressure of the liquid phase equals that of the solid phase. Because the solute lowers vapor pressure, this intersection happens at a lower temperature.

ΔTf = T°f - Tf = Kf × m

Kf = Cryoscopic Constant. m = molality.

4 Thermodynamic Expressions for Kb and Kf

Crucial for JEE/NEET Advanced problems. The constants depend only on the properties of the pure solvent.

Kb = (R × M1 × T°b2) / (1000 × ΔHvap)
Kf = (R × M1 × T°f2) / (1000 × ΔHfus)

R = 8.314 J/(mol·K), M1 = Molar mass of solvent (g/mol), Temperatures must be in Kelvin.

Deep Dive

Want the Complete Detailed Notes?

Read the full comprehensive article for Chapter 1 on our blog, including detailed theory, solved examples, and deep dives for JEE/NEET.

Read Full Article
Test Your Knowledge

Lecture 5 Checkpoint

Answer these quick questions to solidify your understanding of Colligative Properties.

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JEE & NEET Chemistry by Abhishek Sengar

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1 comment:

  1. Anonymous17:03

    Great job explaining this topic.

    ReplyDelete

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