Colligative Properties: Applications, Determination of Molar Mass and Formula

Chemistry Solutions 2025

Jaya Sharma
Updated on Jun 20, 2025 19:11 IST

By Jaya Sharma, Assistant Manager - Content

Colligative properties are physical properties of solutions that depend only on the number of solute particles present, not on their identity. These properties arise due to the disruption of solvent behavior caused by the presence of dissolved substances. NCERT excercise of the chapter 1 has a number of questions based on colligative properties. There are four main colligative properties:

  • Vapor Pressure Lowering
  • Boiling Point Elevation
  • Freezing Point Depression
  • Osmotic Pressure
Table of content
  • List of Different Colligative Properties
  • Conditions for Colligative Properties
  • How To Determine Molar Mass?
  • What are the Different Applications of Colligative Properties?
  • Practice Questions
  • Molar Mass Formula Of Different Colligative Properties
  • Conclusion
View More
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List of Different Colligative Properties

The following properties are governed by the number of solute particles in a dilute solution. IIT JAM and NEET students must learn about these colligative properties in detail to solve conceptual questions:

1. Relative Lowering of Vapor Pressure

Non-volatile solutes reduce the solvent's vapor pressure by occupying surface area, decreasing evaporation. According to Raoult's law: P 0 - P s P 0 = X B = n B n A + n B n B n A   (dilute solution)  

For dilute solutions: P 0 - P s P 0 = w B M A w A M B

Molar mass ( M B  ) can be calculated experimentally using vapor pressure measurements, though less common due to accuracy challenges.

2. Elevation of Boiling Point (Ebullioscopy)

A non-volatile solute lowers the solvent's vapor pressure, requiring a higher temperature to reach atmospheric pressure, thus elevating the boiling point: Δ T b = T b - T b 0 = K b m = K b w B × 1000 M B × w A

where K b  is the molal boiling point elevation constant K k g m o l - 1 . Molar mass: M B = K b w B × 1000 Δ T b w A

 

3. Depression of Freezing Point (Cryoscopy)

A non-volatile solute lowers the vapor pressure, reducing the temperature at which the liquid and solid phases are in equilibrium, depressing the freezing point: Δ T f = T f 0 - T f = K f m = K f w B × 1000 M B × w A

where K f is the molal freezing point depression constant. Molar mass: M B = K f w B × 1000 Δ T f w A

4. Osmotic Pressure (Osmometry)

Osmotic pressure is the pressure required to stop the net flow of solvent molecules across a semi-permeable membrane that separates a solution from pure solvent. This property arises because solvent molecules move from a region of lower solute concentration (or pure solvent) to a region of higher solute concentration, aiming to equalize the chemical potential.

As per the Van't Hoff Equation:
π=CRT
Here:
  • 𝜋 = osmotic pressure (in atm)
  • C = molar concentration of solute (mol/L)
  • R = universal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹)
  • T = temperature in Kelvin
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Conditions for Colligative Properties

The following conditions must be met for colligative properties:

  • Solution must be dilute: Colligative properties work the best for dilute solutions where solute amount is much smaller as compared to the amount of solvent. This interaction of solute-solvent is negligible and solvent behave ideally.
  • Solute must be non-volatile: Solute must not contribute significantly to vapor pressure. This allows for changes in boiling point, freezing point and vapor pressure to be attributed only to solvent. Students of CUET exam must learn about these conditions in detail to perform well in the exam.
  • Solute should not dissociate or associate in solution: Association decreases the number of effective particles. This lowers the colligative effect. Dissociation, on the other hand, increases the number of particles. 
  • Solution must follow Roult's Law: It is important for solution to follow Roult's law. As per this, intermolecular force between solvent and solute should be similar to those in the pure solvent. Along with this, no significant amount of heat is released or absorbed on mixing.
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How To Determine Molar Mass?

Colligative properties are used to calculate the molar mass of unknown solutes:

  • Vapor Pressure: Less preferred due to experimental difficulty.
  • Boiling Point Elevation: Suitable for small solutes; requires accurate K b .
  • Freezing Point Depression: Widely used; K f is solvent-specific.
  • Osmotic Pressure: Best for large molecules (e.g., proteins) due to measurable pressure changes.

Take a Look At These As Well

NCERT Notes

NCERT Class 11 Notes

NCERT Class 11 Chemistry Notes

CBSE Class 12 Chemistry Notes

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What are the Different Applications of Colligative Properties?

The following points highlight the applications of colligative properties:

  1. Chemical analysis: Colligative properties are used for determining molar mass of unknown solutes. Large molecules such as proteins cannot be analyzed easily through conventional means. By measuring properties such as freezing point depression, it is possible to calculate the approximate molar mass. Questions based ojn this application as asked in IISER entrance exam.
  2. Medical Field: Isotonic solutions (e.g., ) In this field, these properties are useful for preparing isotonic solutions that are safe and suitable for biological systems. Take Isotonic Saline  0.9 % N a C l for example. It has the same osmotic pressure as the human plasma. When it is intravenously administered, it prevents the osmotic damage to RBCs which in other condition, could burst or shrink.
  • Industrial Processes: In industrial processes, colligative properties are useful in different industrial setting for safety and temperature regulation. For instance, in colder climates, substance such as ethylene glycol is added to car radiators. This decreases water's freezing point which then prevent ice formation in the negine and damages because of expansion.
  • JEE Main Problems: Numerical calculations of molar mass or colligative property values.
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Practice Questions

  1. A solution of 5 g of an unknown solute in 100 g of water freezes at - 1.86 C . Calculate the molar mass ( K f = 1.86 K k g m o l - 1 ).
  2. Calculate the osmotic pressure of a solution containing 18 g of glucose in 1 L at 293 K R = 0.0821 L K - 1 m o l .
  3. Why is osmotic pressure preferred for determining molar masses of proteins?
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Molar Mass Formula Of Different Colligative Properties

The following table shares formula for molar mass in correpondence to colligative properties:

Property Formula for Molar Mass
Vapor Pressure MB = (wB MA P0) / (wA (P0 - PS))
Boiling Point Elevation MB = (Kb wB × 1000) / (ΔTb wA)
Freezing Point Depression MB = (Kf wB × 1000) / (ΔTf wA)
Osmotic Pressure MB = (wB R T) / (π V)
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Conclusion

Colligative properties are a cornerstone of JEE Main preparation, linking theory to practical applications like molar mass determination. Focus on NCERT formulas, practice numerical problems, and understand real-world contexts like isotonic solutions. Solving JEE-level questions will ensure mastery of these concepts for both exams and beyond.

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