NEET Exam  >  NEET Notes  >  Chemistry Class 12  >  Colligative Properties: Relative Lowering of Vapour Pressure

Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET PDF Download

Did you know that adding salt on icy roads in winter prevents the formation of black ice by lowering the freezing point of water? This ensures safer driving conditions in winter. Do you know how this phenomenon occurs? It's because of a colligative property named "Freezing Point Depression".

What are Colligative Properties?

The colligative properties can be defined as the properties of solutions which are wholly determined by the ratio of the number of solute particles and the number of solvent molecules in a particular solution, and are completely independent of the nature of the chemical species present.

There are four main colligative properties:

  1. Lowering of vapour pressure

  2. Elevation of boiling point

  3. Depression of freezing point

  4. Osmotic pressure

Colligative Properties and Its TypesColligative Properties and Its Types

Applications of Colligative Properties

  • Colligative properties are important in various practical applications, including in the pharmaceutical industry, food science, and chemical engineering
  • They are also used in analytical techniques, such as osmometry, to determine the molecular weight of substances.

Note: It's important to note that colligative properties are only applicable to ideal solutions, which means that the solute-solvent interactions are relatively weak, and the solute particles do not associate or dissociate significantly in the solution. In real-world situations, deviations from ideal behaviour can occur, especially at high solute concentrations or when solute-solvent interactions are strong.

Question for Colligative Properties: Relative Lowering of Vapour Pressure
Try yourself:
What is the colligative property that is responsible for preventing the formation of black ice on icy roads in winter?
View Solution

Relative Lowering of Vapour Pressure

The relative lowering of vapour pressure is one of the colligative properties in chemistry. On adding a nonvolatile solute to a solvent, the vapour pressure of the solution (solute + solvent) becomes lower than the vapour pressure above the pure solvent.

Lowering of Vapour PressureLowering of Vapour Pressure

  • The vapour pressure of a volatile solvent gets lowered when a non-volatile solute is dissolved in it. 
  • If p∘ represents then the vapour pressure of a pure solvent and p represents the vapour pressure of the solution, we have- 
  • Lowering of pressure: po - p
  • Relative lowering of vapour pressure =  po - p/ p
  • The relative lowering of pressure and lowering of pressure are colligative properties. 
  • A relation between the pressure of the solution, the vapour pressure of the pure solvent, and the mole fraction of the solute were discovered by a French chemist Raoult. 
  • Raoult’s law, states that the lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute in the solution.
    Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET
    Here, n is the moles of the solute dissolved in N moles of the solvent.

Vapour Pressure Lowering as a Colligative Property

Vapour pressure is not a colligative property. This is because vapour pressure does not depend on the number of solute particles but depends on the nature of the solute.

  • Relative vapour pressure is a colligative property, i.e. it depends on the number of solute particles. The dissociation of ionic compounds will affect the vapour pressure of the solution. With an increase in the number of dissociated solute ions, the vapour pressure of the solution decreases.
  • For example, it was found that the lowering of vapour pressure of a solution consisting of 1M NaCl is more than that of a solution having 1M  glucose. The concentration of both solutions at standard temperature and pressure is the same. The reason behind this difference can be accounted for by the following reactions.
    NaCl(s)→Na+(aq)+Cl–(aq) ⇒ 2 dissolved particles
    C6H12O6(s)→C6H12O6(aq) ⇒1dissolved particles

Question for Colligative Properties: Relative Lowering of Vapour Pressure
Try yourself:
What is the relative lowering of vapour pressure?
View Solution

  • Sodium chloride is an ionic compound that dissociates into two ions, whereas glucose being an organic compound does not dissociate. Therefore, equal concentrations of sodium chloride and glucose solution will result in twice as many dissolved particles in the case of sodium chloride. The vapour pressure of the sodium chloride solution will be lowered to twice the amount of the glucose solution.

Relative Lowering of Vapour Pressure Formula

In 1886, the French chemist, Francois Raoult, after a series of experiments on several solvents including water, benzene, and ether, succeeded in establishing a relationship between the lowering of the vapour pressure of a solution and the mole fraction of the non-volatile solute.

Experimentally, we know that the vapour pressure of the solvent above a solution containing a nonvolatile solute (i.e., a solute that does not have a vapour pressure of its own) is directly proportional to the mole fraction of solvent in the solution. This behaviour is summed up in Raoult’s Law:

Psolution = Xsolvent × Posolvent
In a binary solution, if a volatile solute is added to the volatile solvent, each solute’s and solvent’s component is added to the total pressure
Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

According to Raoult’s Law, the partial vapour pressure of two components, solvent (A) and solute (B), of a solution may be given as:

Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEETColligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

Vapour Pressure Lowering Example Problems With Solutions

Q.1: At 25oC the vapor pressure of pure benzene is 93.9 torr. When a nonvolatile solute is added to benzene, the vapour pressure of benzene is lowered to 91.5 torr. Calculate the mole fraction of the solute and the solvent.

Ans: 
Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

Q.2: The vapour pressure of a pure liquid at 298K is 4×104Nm2. On adding a nonvolatile solute, the vapour pressure of the solution becomes 3.65×104Nm2. Calculate the relative vapour pressure, lowering of vapour pressure, and relative lowering of vapour pressure.

Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

Q.3: The density of a 0.438 M solution of potassium chromate at 298 K is 1.063 g cm-3. Calculate the vapour pressure of water above this solution. Given: Po (water) = 23.79 mm Hg.
Solution. 
A solution of 0.438 M means 0.438 mol of K2CrO4 is present in 1L of the solution. Now,
Mass of K2CrO4 dissolved per litre of the solution  = 0.438 × 194  = 84.972 g
Mass of 1L of solution = 1000 × 1.063 = 1063 g
Amount of water in 1L of solution = 978.028/18  = 54.255 mol
Assuming K2CrO4 to be completely dissociated in the solution, we will have;
Amount of total solute species in the solution = 3 × 0.438 = 1.314 mol.
Mole fraction of water solution = 54.335/(54.335 + 1.314) = 0.976
Finally, Vapour pressure of water above solution = 0.976 × 23.79  = 23.22 mm Hg

Determination of Molar Masses from Lowering of Vapour Pressure

  • It is possible to calculate molar masses of non-volatile non-electrolytic solutes by measuring the vapour pressures of their dilute solutions.
  • Suppose, a given mass, w gram, of a solute of molar mass m, dissolved in
    W gram of solvent of molar mass M lowers the vapour pressure from Po1 to P1.
  • We know that the relative lowering of the vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute present in the solution.
  • This implies
    Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

Measurement of Lowering of Vapour Pressure

  1. Barometric Method: 
    Raoult initially devised a method to determine the individual vapor pressure of a liquid and subsequently applied the same approach to calculate the vapor pressure of a solution. This involved introducing either the liquid or the solution into a Torricellian vacuum within a barometer tube, enabling the measurement of the depression in the mercury level. However, it was later discovered that this method proved impractical and inaccurate, primarily due to the minimal reduction in vapor pressure observed.
  2. Manometric Method: 
    To accurately measure the vapor pressure of a liquid or solution, a manometer proves to be a reliable tool. In this method, a bulb is filled with the liquid or solution, and the air in the connecting tube is evacuated using a vacuum pump. By closing the stopcock, the pressure inside the apparatus is solely attributed to the vapor emanating from the solution or liquid. This technique applies to aqueous solutions, and the manometric liquid employed can either be mercury or n-butyl phthalate, chosen for their low density and volatility.
  3. Ostwald and Walker’s Dynamic Method (Gas Saturation Method):
  • When air is introduced into a gas, the gas diffuses into the air following the principle of diffusion until the pressure of the gas in the air matches that outside the gas. Consequently, as air passes through a solution, it takes in the vapors of the solvent until the pressure of the solvent vapors in the air reaches P (the vapor pressure of the solvent in the solution). 
  • The mass of vapor absorbed or the loss in weight of the solution can be expressed as PVm/RT, where P is the pressure of the solvent vapors, V is the volume available for vapors in the solution bulb, T is the temperature of the bulb, and m is the molecular weight of the solvent. As the air traverses the solvent, the pressure difference between the vapors in the air and outside is Po - P, leading to the air saturating with this amount of vapors. The loss in weight of the solvent bulb is given by (Po - P)Vm/RT. The combined loss in weight of the two containers equals the gain in weight of CaCl2.
    The gain in weight of CaCl2-filled tubes =
    Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET
    Method: In this method, a stream of dry air is passed successively through
    (i) a solution
    (ii) the pure solvent (water) and
    (iii) a reagent (anhydrous CaCl2) that can absorb the vapors of the solvent. The complete assembly is shown in the figure, given below:

Ostwald Walker ExperimentOstwald Walker Experiment

  • The first three bulbs contain a weighed amount of the solution under examination and the next three bulbs contain a weighed amount of the pure solvent.
  • A weighed amount of CaCl2 is taken in the set of U-tubes at the end. All the bulbs are at the same temperature and the volume available for the vapors in the solution and solvent bulbs are the same temperatures and the volume available for the vapors in the solution and solvent bulbs are the same.

It is designed to calculate the relative lowering of the vapor pressure of a solvent due to a non-volatile solute.

Some Important Questions

Q.1. What are the properties arising due to varying concentrations of solute in a given solvent, irrespective of the nature of the solute concerning the solvent?

a) Colligative properties
b) Intensive properties
c) Extensive properties
d) Solute properties
Ans:  a
 Colligative properties are a set of four properties. This set of properties purely depends on the number of solute particles present in the solution/solvent, independent of the nature of the particles concerning the solution/solvent.
The properties are
1.) Relative lowering of vapor pressure,
2.) Elevation in boiling point,
3.) Depression in freezing point and
4.) Osmotic pressure.

Q.2. Which law specifically governs the relative lowering of vapor pressures in solutions?
a) Van’t Hoff law
b) Boyle’s law
c) Raoult’s law
d) Amagat’s law

Ans:  c
 Raoult’s law quantifies the relative lowering in vapor pressure. From the law, it follows that p1 = X1 x po1. If p01 was the original pressure before the X2 mole fraction of solute was added to the solvent then a reduction in vapor pressure is given as:

Δp1 = po1 – p

= po1 – po1 X1

= po1(1 – X1)

= po1X2

Therefore, the relative reduction in vapor pressure (∆p1/po1) = X2 i.e. mole fraction of solute in the solution.

Q.3. On addition of non-volatile potassium iodide in water at 298K it is noticed that vapor pressure reduces from 23.8 mm Hg to 2.0 cm Hg. What is the mole fraction of solute in the solution?
a) 0.916
b) 0.160
c) 0.084
d) 0.092

Ans:  b
 Given,
P0water =23.8 mm Hg
Pwater= 2.0 cm Hg = 20.0 mm Hg (after addition of solute)
From the law of relative lowering of vapor pressure, ∆p1 = X2 x po1 (where X2 is the mole fraction of solute)
On rearranging, Δp1/po1 =X2
Δp1 = 23.8 – 20.0 = 3.8 mm Hg
X2 = 3.8/23.8 = 0.160

Q.4. Which of the following is Raoult’s law applicable to, to determine molar masses correctly?

a) Ionic solute in liquid
b) Non-ionic solute in dilute solution
c) Non-ionic solute in concentrated solution
d) Ionic solid in insoluble form in solvent

Ans:  b
 To determine molar masses correctly, the solute must be non-volatile, non-ionic, and present in dilute form only. If it is ionic one has to account for its van’t Hoff factor, i, for association in concentrated solution and dissociation in dilute solution. Molar masses can only be calculated from dilute solutions containing non-dissociable non-ionic solutes. 

Q.5. Which of the following is a colligative property?

a) Relative lowering of fluid pressure
b) Decrease in boiling point
c) Decrease in freezing point
d) Change in volume after mixing

Ans:  c

 A decrease in freezing point is the correct colligative property, known as ‘Depression in freezing point’. This is caused by solute particles present on the surface which lowers the equilibrium solid-vapor pressure. Therefore, a lower freezing temperature is required to match the pressure outside. The other correct colligative properties are
1.) Relative lowering of vapor pressure,
2.) Elevation in boiling point,
3.) Depression in freezing point and
4.) Osmotic pressure.

Q.6. Mass of Urea (NH2CONH2) required to be dissolved in 1000 g of water in order to reduce the vapour pressure of water by 25% is _________ g. (Nearest integer) 
Given: Molar mass of N, C, O and H are 14,12,16 and 1 g mol−1 respectively
Ans: 1111
Given: Vapor pressure reduction: 25% (0.75 times the vapor pressure of pure water)
Molar mass of water (H2O): 18g/mol
Mass of solvent (water): 1000g
Using Raoult's law:
Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

Solving for (x):

Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET
So, the mass of urea required to be dissolved in 1000g of water is 1111g.

The document Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET is a part of the NEET Course Chemistry Class 12.
All you need of NEET at this link: NEET
150 videos|378 docs|213 tests

FAQs on Colligative Properties: Relative Lowering of Vapour Pressure - Chemistry Class 12 - NEET

1. What are colligative properties?
Colligative properties are physical properties of a solution that depend on the concentration of solute particles, rather than the nature of the solute itself.
2. What is the relative lowering of vapor pressure?
The relative lowering of vapor pressure is a colligative property that describes the decrease in vapor pressure of a solvent when a non-volatile solute is added to it. It is defined as the ratio of the change in vapor pressure of the solvent to the vapor pressure of the pure solvent.
3. What is the formula for relative lowering of vapor pressure?
The formula for relative lowering of vapor pressure is: Relative Lowering of Vapor Pressure = (Change in Vapor Pressure of Solvent) / (Vapor Pressure of Pure Solvent)
4. Can you provide an example problem with solution for vapor pressure lowering?
Example problem: Calculate the relative lowering of vapor pressure when 10 grams of sucrose (molar mass = 342 g/mol) is dissolved in 250 grams of water. The vapor pressure of pure water at a certain temperature is 23.8 mmHg. Solution: Step 1: Calculate the number of moles of sucrose: Number of moles = mass of sucrose / molar mass of sucrose Number of moles = 10 g / 342 g/mol = 0.0292 mol Step 2: Calculate the molality of the solution: Molality = moles of solute / mass of solvent (in kg) Molality = 0.0292 mol / 0.250 kg = 0.1168 mol/kg Step 3: Calculate the change in vapor pressure of water: Change in vapor pressure = Molality * Constant (for water, the constant is approximately 0.52) Change in vapor pressure = 0.1168 mol/kg * 0.52 = 0.0607 atm Step 4: Calculate the relative lowering of vapor pressure: Relative Lowering of Vapor Pressure = Change in Vapor Pressure / Vapor Pressure of Pure Solvent Relative Lowering of Vapor Pressure = 0.0607 atm / 1 atm = 0.0607 Therefore, the relative lowering of vapor pressure is 0.0607.
5. How is the lowering of vapor pressure measured?
The lowering of vapor pressure can be measured using various methods, such as the boiling point elevation method or the freezing point depression method. These methods rely on the fact that the vapor pressure of a solution is lower than that of the pure solvent, and by measuring the temperature at which the solution boils or freezes, the lowering of vapor pressure can be determined.
150 videos|378 docs|213 tests
Download as PDF
Explore Courses for NEET exam

How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches

Extra Questions

,

Summary

,

mock tests for examination

,

video lectures

,

shortcuts and tricks

,

Sample Paper

,

Exam

,

ppt

,

Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

,

Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

,

Viva Questions

,

Objective type Questions

,

practice quizzes

,

Important questions

,

Previous Year Questions with Solutions

,

Free

,

past year papers

,

MCQs

,

Semester Notes

,

study material

,

Colligative Properties: Relative Lowering of Vapour Pressure | Chemistry Class 12 - NEET

,

pdf

;