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Colligative properties refer to those properties which are entirely dependent on the ratio of the number of solute particles to the total number of solvent particles. The chemical nature of the particles of the solution does not affect colligative properties. In this article, we will cover the definition of colligative properties, types, Van’t Hoff factor, and previous year sample questions. Check CBSE Class 12 Chemistry Syllabus.
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Key Terms: Colligative properties, Dilute solutions, Elevation in boiling point, Raoult’s Law, Van’t Hoff factor.
Definition of Colligative Properties
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Colligative properties are the properties that help us to understand how a solutions’ properties are linked to a solute’s concentration in any solution. These properties are only dependent on the number of non-volatile solute particles and not on their type. Colligative properties are widely observed in dilute solutions.
We can also consider colligative properties as the properties observed by the dissolving of a non-volatile solute in an unstable solvent. Normally, the solvent properties are changed by the solute where its particles eliminate a portion of the solvent atoms in the fluid stage. This likewise brings about the decrease of the grouping of the solvent.
Colligative properties can be linked with multiple quantities which are useful in expressing the concentration of a solution, such as molarity, normality, and molality. The various types of colligative properties are:
- Freezing point depression
- Boiling point elevation
- Osmotic pressure
- Relative lowering of vapor pressure
The word ‘colligative’ is derived from the Latin word Colligatus. Colligatus translates to ‘bound together.’ In the situation of defining a solution, Colligative properties help us understand how the properties of the solution are associated with the concentration of solute in the solution.
The video below explains this:
Colligative Properties and Determination of Molar Mass Detailed Video Explanation:
Also Read:
| Related Articles | ||
|---|---|---|
| Solutions | Dalton's Law of Partial Pressure | Solubility |
| Abnormal Molar Masses | Ideal and Non-ideal Solutions | Ideal Solution |
Examples of Colligative Properties
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We can notice the colligative properties of arrangements by going through the accompanying examples:
- On the off chance that we add a spot of salt to a glass full of water, its freezing temperature is brought down impressively than its normal temperature. On the other hand, the boiling temperature is likewise increased and the arrangement will have a lower vapor pressure. There are also changes observed in its osmotic pressure.
- In the same way, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is usually observed for either pure alcohol or water.
Different Types of Colligative Properties
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There are many types of colligative properties of a solution. Let us discuss them in detail:
Elevation in Boiling Point
The boiling point of water can be understood as the temperature at which the vapour pressure of water is the same as the atmospheric pressure. We also know that when a pure solvent is added to a non-volatile liquid, the solution’s vapour pressure decreases. Hence to make vapour pressure and atmospheric pressure the same, we have to increase the temperature of the solution. This difference between the solution’s boiling point and the boiling point of a pure solvent is known as the elevation in boiling point.
Elevation in Boiling Point
ΔTb =T0b-Tb
where, ΔTb: Elevation in the Boiling Point
T0b: Boiling Point of a Pure Solvent
Tb: Boiling point of a Solution
Experimentally, it is observed that there’s a correlation between the elevation in boiling point and the molality of the solute present in the solution represented by ‘m’.
ΔTb ∝ m
ΔTb = kb m
where, kb: molal elevation constant
Hence, by substitution of the value of ‘m’
ΔTb =1000 x kb x m2 /M2 x m1
where, m2: mass of solvent (g)
m1: mass of solvent (kg)
M2: Solute’s molar mass
Lowering of Vapour Pressure
A volatile solvent’s vapor pressure is lowered when a non-volatile solute is dissolved in it. Due to the presence of both solute and solvent particles on the surface of the solution, the surface area covered by the molecules of the solvent is reduced. Hence, the vapour pressure is decreased as it solely depends on the solvent.
Now, Lowering of pressure = Po–Ps
And Relative lowering of vapor pressure = (Po - Ps)/Po
where, Po: Vapor pressure of the pure solvent
Ps: Vapor pressure of the solution
Raoult’s law
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In 1886, Raoult derived a relation between the mole fraction and relative lowering in vapour pressure.
Pa = XAPoA
It’s given as - (Po – Ps)/ Po = n/(n + N)
Where in n moles of solvent, n moles of solute are dissolved.
Depression in Freezing Point
The temperature at which a liquid’s vapour pressure is equivalent to the vapour of the corresponding solid is known as the Freezing Point. In accordance with Raoult's law, the vapour pressure of a solvent decreases when a non-volatile solid is added to it and becomes equal to the vapor pressure of the solvent at a lower temperature. Hence, this difference between the freezing point of a pure solvent and the freezing point of its solution is known as the depression in the freezing point.
ΔTf = T0f – Tf
ΔTf = – kf.M
where, ΔTf: Depression in the Freezing Point
Tof: Boiling Point of a Pure Solvent
Tf: Boiling point of a Solution
Like the boiling point, the freezing point also directly relates to the molality.
ΔTf =1000 x kf x m2 /(M2 x m1)
Where kf = molal depression constant
m2 = mass of solvent (g)
m1 = mass of solvent (kg)
M2 = Solute’s molar mass
Osmotic Pressure
The solvent particles enter the solution when a semipermeable membrane is placed between a solvent and solution, increasing the volume of the solution. The semi-penetrable membrane permits just the solvent particles to go through it preventing the entry of larger atoms like solute. This phenomenon wherein the molecules of solvent flow spontaneously from a dilute to a concentrated solution or from a pure solvent to the solution is called Osmotic Pressure.
Osmotic Pressure is a colligative property as it is not based on the nature of the solute but on the number of solute particles present. It has been proved experimentally that osmotic pressure is directly proportional to the temperature (T) and molarity(C).
Hence, π = CRT
where, R: Gas constant.
⇒ π = (n2/V) RT
where there are n2 moles of solute and
V: Volume of the solution
=> n2= m2/M2
m2: weight of the solute
M2: Molar mass of solute
Thus, π = W2RT / M2V
Van’t Hoff Factor of Colligative Properties
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When the solute is dissolved in a solution, it undergoes either association or dissociation. Hence, the colligative properties change as the number of particles increases or decreases. Van’t Hoff Factor can be used to express the extent of this association or dissociation.
i = Normal molar massAbnormal molar mass
i = Calculated Colligative Property Observed Colligative Property
Things to Remember
- Colligative properties refer to those properties which are entirely dependent on the ratio of the number of solute particles to the total number of solvent particles.
- The various types of colligative properties of different solutions are:
- Freezing point depression: ΔTf =1000 x kf x m2 /(M2 x m1)
- Boiling point elevation: ΔTb = kb m
- Osmotic pressure: π = (n2/V) RT
- Relative lowering of vapor pressure: (Po - Ps)/Po
- Van’t Hoff Factor, i = observed colligative property/calculated colligative property.
- Read more about different types of solutions here.
Also Read:
Previous Years Questions
- what is the molecular weight of the solute?...[BITSAT 2008]
- The vapour pressure of due to….[KCET 2012]
- Find out the molecular weight of the substance...[NEET 1999]
- 1 M and 2.5 litre NaOH solution mixed with another….[NEET 1980]
- an aqueous solution of a protein contains its….[NEET 2011]
- A 0.1 molal aqueous solution of a weak acid is...[NEET 2011]
- For an ideal solution, the correct option is….[NEET 2019]
- the highest osmotic pressure is exhibited by 0.1 M solution of...[NEET 1994]
- Which one of the following salt will have the same value of vant Hoff's...[NEET 1994]
- Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be...[NEET 2009]
- A 5% solution of cane sugar (MW = 342) isisotonic with 1%….[NEET 1998]
- Pure water can be obtained from sea water by...[NEET 2001]
- the boiling point of this solution will be….[NEET 2016]
- The molecular mass of this non-volatile solute is….[NEET 2006]
- A solution of acetone in ethanol…. [NEET 2006]
- Which change will cause the vapour pressure of the solution to increase?...[NEET 2010]
- An ideal solution is formed when its components….. [NEET 1988]
- For determination of molar mass of colloids,polymers and protein which colligative property is used….[NEET 2000]
Sample Questions
Ques. Estimate the osmotic pressure of a 5% solution of urea at 273K. (Mol.Mass=60)(R=0.0821LatmK–1mole) (3 marks)
Ans. 5% solution of urea means that it contains 5g of urea per 100cm3 of the solution, i.e.,
W2 = 5g,
V=100cm3 = 0.1L
M2 = 60gmol–1,
R= 0.0821LatmK–1mol–1,
T= 273K
π = W2.RT/M2V
= 5g × 0.0821LatmK–1mol–1 × 273K / (60gmol–1×0.1L)
=18.68atm.
The osmotic pressure is 18.68atm.
Ques. A solution has 0.456g of camphor (mol. mass=152) in 31.4g of acetone (boiling point:56.30oC). Calculate its boiling point if the molecular elevation of acetone is 17.2oC per 100g. (5 marks)
Ans. m2=0.456g,
M2 =152 ,
m1=31.4g,
To=56.30oC,
kb=17.2oC/100gm
Hence, kb per 1000 gm of acetone: 1.72oC
ΔTb = kb m
=1000* kb* m2 / M2*m1
=1000 × 1.72 × 0.456/(152 x 31.4)
Therefore, the Boiling point of the solution (Tb)= Tbo+ΔTb
= 56.30+0.16
=56.46?C.
Ques. When 1g of non-electrolyte solute dissolves in a given 50g of benzene, it lowers its freezing point by 0.40K. The freezing depression constant of benzene is given as 5.12Kkgmol–1. Calculate the molar mass of the solute. (2 marks)
Ans. Given w2 = 1g,
m1=50g,
ΔTf=0.40K,
kf= 5.12Kkgmol–1
Putting these values in the formula
ΔTf =1000 x kf x m2 /M2 x m1
M2=1000 x kf x m2/ m1 x ΔTf
M2= 1000 × 5.12 / (50 × 0.40)
=256gmol–1.
Ques. The molal elevation stable for liquid is 0.513oC kg mol–. While 0.2mole of sugar is dissolved in 250g of liquid, compute the temperature at which the solution boils under atmospheric pressure. (2 marks)
Ans. ΔTb = moles of sugar x 1000/weight of water in gm
ΔTb = 0.2 x 1000 / 250
ΔTb = 0.8
⇒ T0b-Tb = 0.8
For pure liquid, T0b =100oC
⇒ Tb= 0.8 + 100
=100.80oC
Ques. 300 cm3 of an aqueous solution contains 1.56g of a polymer. The osmotic pressure of such solution at 300 K is set up to be 2.57 ? 10-3 bar. Find out the molar mass of the polymer. (3 marks)
Ans. Weight of polymer( W2) = 1.56g
According to the value of Osmotic pressure (π) = 2.57 ? 10-3 bar
Volume (V) = 300 cm3 = 0.3L
M2 = W2 RT / π V
Putting all values according to the formula
M2 = 1.56 x 0.083 x 300 / (0.3 x 2.57 x 10-3)
= 50381 g mol–
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