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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Chemistry » Lowering of Vapour Pressure

Lowering of Vapour Pressure

The article discusses the colligative properties of solutes and solvents, their equation and focuses on lowering vapour pressure.

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  • There are two types of physical properties, extensive properties that deal with mass and volume, and intensive properties that deal with concentration and are independent of the size and sample being studied. 

  • Apart from this, there is the third property, which is in between extensive and intensive property, known as colligative property. 

  • Properties that are dependent on the ratio between the number of solvent particles and the number of solute particles are called colligative properties. 

  • The term colligative comes from a Latin word that means “to bound”. These properties can be expressed in quantities describing concentrations like molality, normality, molarity, and do not depend on the components present in the solution. 

  • Among the four colligative properties, that is, osmotic pressure, elevation in boiling point, depression in freezing point, and lowering of vapour pressure, the article will discuss the colligative property that deals with the lowering of vapour pressure.

  • Lowering of vapour pressure, boiling factor elevation, freezing factor depression, and osmotic stress are all examples of collaborative properties. 

  • This little cluster of dwellings has significant implications for a variety of herbal phenomena and technology applications, as explained in this module.

  • Consider a beaker containing solvent whose surface is occupied by a pure solvent molecule. Once you add non-volatile solute to the above solvent, the surface of the beaker is now filled with solute, as well as solvent, eventually decreasing the solvent molecules present on the surface. 

  • In this case, there was a lowering of vapour pressure because of the addition of solute particles, and no other factors such as temperature and concentration affected the vapour pressure. In this case, Po is the vapour pressure of the solvent and Ps is the vapour pressure of the solution. 

  • The ratio or the difference between the vapour pressure of the solvent and solution will give us a lowering of vapour pressure and the ratio Po– Ps / Po is called the relative lowering of vapour pressure.

Raoult Law

In the year 1886, Raoult came up with a relationship between the relative lowering of vapour pressure and mole fraction and termed it Raoult’s law. This law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute in the solution by the relation:

(Po– Ps )/ Po = n / n+N

Here,

No. of moles of solute = n

No. of moles of solvent = N

Vapour pressure of pure solvent = p0

Vapour pressure of solution= Ps

Elevation in boiling point: If a non-volatile solute is introduced to a solvent, the boiling point of the solvent is raised. In such a case, the molal concentration of the solute added to the solution and the rise in boiling point is directly proportional to each other. Such a phenomenon is called elevation in boiling point.

Consider the equation:

ΔTb = Kbm = (1000 w2) (w1 M2) 

Due to the increase in boiling point, the results are indicated by:

ΔTb = (Kb × 1000 × w2) / (w1 × M2)

In the above equation, the molecular weight of solute becomes:

M2 = (Kb × 1000 × w2) / (w1 × △Tb).

Here, 

w1 = mass of solvent

w2=mass of solute

M2=molar mass of solute 

m = molality

Kb is a constant called Boiling Point Elevation Constant.

Osmotic Pressure: The pressure necessary to prevent water from diffusing across a barrier created via osmosis is called osmotic pressure. The diffusion of water across a semipermeable membrane is known as osmosis. As a result, the solutes are unable to move across the semipermeable membrane as they are unable to pass.

Consider the following example:

π=CRT

Where,

π = is denoted as osmotic pressure.

C= Molar concentration of the solution (The number of atoms, ions, or molecules in a solute is measured in molar concentration.)

R= Universal gas constant.

T= Temperature in degrees or Kelvin.

Depression in freezing point: When a non-volatile solute is introduced into a solvent, the freezing point of the solvent is lowered. This phenomenon is called Depression in freezing point. For example,

  • Addition of salt in water.

  • Mixing water with alcohol. 

The freezing point of the resulting solution or combination is lower than that of a pure solvent. The molal concentration of the solution’s solute and the decrease in freezing point are directly proportional. 

This decrease in the freezing point is expressed by the equation:

ΔTf = Kf × m.

Tf is the freezing point depression in this equation.

Conclusion

While different properties have different characteristics, the article focuses on lowering vapour pressure. It’s quite evident from Raoult’s law that the mole fraction of solutes is related to the lowering of vapour pressure. Also, the number of solute particles present in the solution lowers the relative vapour pressure of the solvent. It was also evident that the concentration of solute/solvent does not play a major role in lowering the vapour pressure.

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

Which factors are to be considered to determine molar mass?

Ans.  There is a shift in boiling point...Read full

Why can’t solids move when osmotic pressure is applied?

Ans. The pressure used to halt the passage of solute molecules from a dilute solution to a concentrated solution ove...Read full

State Raoult's law of lowering vapour pressure?

Ans. In the year 1886, Raoult came up with a relationship between the relative lowering of vapour pressure and mole fraction and termed it Raoult&#...Read full

Lowering of vapour pressure depends on which factors?

Ans. Lowering of vapour pressure depends only on the no. of solute particles present in the solution and is independent of other factors like conce...Read full

Ans.

  1.  There is a shift in boiling point: T = Tsolution – Tpure solvent
  2. Divide ΔT by the constant boiling point elevation to get the molal concentration.
  3. Finally, given the mass of an unknown number of moles, calculate the molar mass. Divide the unknown’s mass by the unknown’s moles to arrive at this result.

Similarly, the freezing point depression and the osmotic pressure may be used to calculate molar mass. 

 

Ans. The pressure used to halt the passage of solute molecules from a dilute solution to a concentrated solution over a semi-permeable barrier is known as osmotic pressure. Here, the size of the membrane is so small that only solvent can pass through it freely.

Ans. In the year 1886, Raoult came up with a relationship between the relative lowering of vapour pressure and mole fraction and termed it Raoult’s law. This law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute in the solution by the relation: Po– Ps Po = n / n+N

Ans. Lowering of vapour pressure depends only on the no. of solute particles present in the solution and is independent of other factors like concentration, size, density, etc. According to Raoult’s law discovered in 1886, lowering of vapour pressure and mole fraction are interrelated systems.

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