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Colligative Properties and Determination of Molar Mass

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Chemists were unable to calculate the molar mass of a solute in a solution and to calculate this molar mass they introduced the terms called colligative properties and there are four colligative properties are Relative lowering in vapour pressure, elevation in boiling point, depression in freezing point, osmotic pressure and to study this we need to know some standard terms and standard definitions.

Molarity (M) 

Molarity is the concentration of a solution measured as a number of moles of solute per litre of solution.         

Molarity = number of moles of solute/volume of solution in L. 

Molality(m)

Molality is the amount of a substance dissolved in a certain mass of solvent. It is defined as the number of moles of a solute per kilograms of a solvent. 

Molality = number of moles of solute/weight of solvent in Kg.

Mole fraction(x)

The mole fraction of a solute in a solution gives the ratio of the number of moles of the solute present in the solution to the total number of moles of the solute and the solvent present in the solution.

There is another definition for mole fraction that is, the mole fraction of a compound is the ratio of the number of moles of the compound to the total moles of compounds in the mixture.      

Mole fraction = number of moles of solute/total number of moles of solute and solvent.

Van’t Hoff factor(i)

The Van’t Hoff factor is the effect of solutes on the colligative properties of solutions. It is denoted by the symbol (I), i is the ratio of the concentration of particles formed when a substance is dissolved to the concentration of the substance by mass, i is the ratio of the theoretical molecular weight and observed molecular weight.
i is the ratio of the final number of particles and the initial number of particles, i is the ratio of observed colligative property and theoretical colligative property.

And, i =1 + α(n – 1) 

here, n is the number of ions formed from one formula unit of a substance, 

α is the degree of dissociation, 

and based on it we have 3 types of electrolytes, Strong Electrolytes, weak electrolytes, non-electrolytes 

  • Strong electrolytes

The electrolytes which completely ionize in water are strong electrolytes. This means 100% of the dissolved chemical divides into cations and anions. example, HCl, NaOH, NaCl. 

  • Weak Electrolytes

Electrolytes that partially ionize in water are weak electrolytes. The electrolytes which dissociate into ions between 0% and 100% is a weak electrolyte. example, CH3COOH, NH

  • Nonelectrolytes

A substance that doesn’t ionize in water is a nonelectrolyte. Examples: CH4, sugars like glucose-fructose, urea.

  • The universal gas constant(R)

It is equal to the product of the pressure and the volume of one mole of an ideal gas divided by the absolute temperature. R has different values for different units, for example, R = 8.314 Jmol-1K-1 = 0.0821 Latmmol-1k-1

  • Temperature 

Temperature is the measure of hotness or coldness expressed in terms of any of several scales, like Fahrenheit, Celsius, kelvin, etc Temperature indicates the direction in which heat energy will spontaneously flow i.e., from a hotter body (one at a higher temperature) to a colder body (one at a lower temperature). 

  • Partial pressure

Partial pressure is the pressure that an individual gas exerts in a mixture of gases and this is equal to the product of mole fraction of a gas and total pressure exerted on the container.

Colligative properties

A binary solution is a mixture of two liquids that are completely miscible one with another, and there are two types of binary solutions,

  1. Both solute and solvent are volatile
  2. The solute is nonvolatile and the solvent is volatile

If both solute and solvent are volatile then,

PTotal = PA + PB (Here PA is partial pressure of A gas in air and PB is partial pressure of B gas in air)

From Raoults law: PA = PA0 x        

Here P0 is the Partial pressure of A in the container.

So PTotal = PA0xA + P0Bx

Here, xB = 1 – xA and viceversa can happen. 

If Solute is nonvolatile and the solvent is volatile

Let’s assume  B as a nonvolatile solute  

So PB0 is 0 so PB is also 0 so,  

PTotal = PA = PA0xA = PA0 (1 – xB

So xB = PA0 – PT/PA0  

Hence, ΔP/P0A = xB   

Here, P0A – PT/PoA is relative lowering of vapor pressure. 
 

  • Relative Lowering in vapour pressure

The relative lowering of vapour pressure is the ratio of lowering of vapour pressure to the vapour pressure of the pure solvent. Therefore, the relative lowering of the vapour pressure of a solution having non-volatile solutes is equal to the mole fraction of the solute.

  • Elevation in boiling point (ΔTb)

Boiling point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. 

ΔTb = TIb – T

ΔTb is  proportional to the number of particles

ΔTb is  proportional to i×m 

ΔTb = I × kb × m (i depends on the type of solute and kb depends on the type of solvent.) 

  • Depression in freezing point (ΔTf)

Freezing Point depression. The freezing point of a solution is less than the freezing point of the pure solvent. This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur.

ΔTf = Tf – TfI 

ΔTf  is proportional to the number of particles

ΔTf  is proportional to I × m

ΔTf = i × kf × m (i depends on the type of solute and kb depends on the type of solvent.) 

  • Osmotic pressure (π) 

Osmotic pressure is the minimum pressure that needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in a pure solvent by osmosis. 

 π = iCRT 

Here i is van’t Hoffs factor, C is concentration (or) molarity, R is the universal gas constant, T is temperature.

In an isotonic solution, the flow of water in and out of the cell is happening at the same rate. Water moves into and out of cells by osmosis.
If a cell is in a hypertonic solution, the solution has a lower water concentration than the cell cytosol, and water moves out of the cell until both solutions are isotonic.
If a cell is in a hypotonic solution, the solution has a high water concentration than the cell cytosol, and water moves into the cell until both solutions are isotonic.  

Determination of molar mass from colligative properties

  • Determination of molar mass from Relative lowering in vapour pressure:

As ΔP/P°A = xB   

As xB = nB/nA + nB  

We know that number of moles = mass/molar mass

From this molar mass is obtained.

  • Determination of molar mass from elevation in boiling point and depression in freezing point:

As ΔTb = i × kb × m and ΔTf = i × kf × m

As Molality = number of moles of solute/weight of solvent in Kg 

As the number of moles = mass/molar mass 

From this, the molar mass is obtained.

  • Determination of molar mass from osmotic pressure:

As π = iCRT 

Where C = molarity

Molarity = number of moles of solute/volume of solution in L 

As the number of moles = mass/molar mass

From this, the molar mass is obtained.                             

Sample Problems

Question 1: Find the relative lowering of vapour pressure, if 18 g of glucose is dissolved in 90 g of water.

Solution: 

Mass of glucose = 18g, 

number of moles of glucose(nB) = 18/180 = 0.1 

Mass of water = 90g, number of moles of water(nA) = 90/18 = 5 

Relative lowering ΔP/PA° is equal to XB.

The value of XB = nB/nA + nB  

So XB = 0.1/0.1 + 5 = 1/51  

Question 2: The boiling point elevation of a solution containing sucrose and water is 0.256c. The molal elevation constant of water is 0.512c/m and the molar mass of sucrose is 342 g/mol. What is molality?

Solution:

Given: Boiling point elevation(ΔTb) = 0.256

molal elevation constant(Kb) = 0.512 

As (ΔTb) = i × Kb × m = Kb × m (Fora nonvolatile solute i is 1 as α is 0)

0.256 = 0.512 × m

From this, m = 0.5mole/kg       

Question 3: Calculate the freezing point depression and the freezing point after adding 100.0 g of table salt to 400.0 g of water. (Kf of water = 1.86) 

Solution:

Moles of NaCl = mass/molar mass = 100.0/58.443 = 1.71107 mol

Mass of water = 400.0 g = 0.400 kg 

Molalilty(m) = moles of NaCl/mass of water in kg = 1.71107/0.400 = 4.2777m

NaCl ⇢  Na+ + Cl

Van’tHofffactor(i) = Numberof mole after dissociation/numberof mole before dissociation 

= 2 (As each NaCl unit fully ionizes into 2 ions)

Freezing point depression constant for water Kf = 1.86

Freezing point depression = i × Kf × m = 2 × 1.86 × 4.2777 = 15.90C

Freezing point of solution = freezing point of water – freezing point depression = 0.0 – 15.9 = -15.90C

Question 4: If 6.8% w/v of cane sugar is isotonic with 1.52% w/v with Thiocarbamide if the molecular weight of cane sugar is 342 find the molecular weight of Thiocarbamide?

Solution:

In an isotonic solution, the flow of water in and out of the cell is happening at the same rate, i.e;

 π1 = π2  i1C1RT = i2C2RT  

From this, i1C1 = i2C

Here, both canesugar and thiocarbamide are non electrolytes,

So i = 1, So C1 = C2 

% (W/V) percent  is the number of grams of solute in 100 mL of solution.

 And C is molarity and Molarity = number of moles of solute/volume of solution in L

 So C1 = 6.8 × 1000/342 × 100 and C2 = 1.52 × 1000/x × 100

 And as C1 = C2  

By solving these we get x = 76

Question5: The osmotic pressure, of a solution of glucose, is 117.4 atm. Find the molarity of the solution at 298 K.

Solution:

π = iCRT    

(glucose is a non electrolyte so i = 1)

117.4 = C × 0.0821 × 298 from this, C = 4.8 mole/lit   

Question6: x grams of solute is dissolved in 500gram solvent if the sum of elevation of boiling point and depression in freezing point for sucrose in water is 5  find its molality. (if Kb = 0.52 and Kf = 1.86) find x?

Solution:

ΔTf + ΔTb = 5 (ΔTf = i × kf × m and ΔTb = i × kb × m)

And, here i = 1 for nonelectrolyte like sucrose.

kf × m + kb × m = 5 

⇒ 2.38 × m = 5

m = 2.1mole/kg 

(Molality = number of moles of solute/weight of solvent in Kg)

 x × 1000/342 × 500 = 2.1 

⇒  x = 359

Question7: If in a container there are 2 gases in which one has a concentration of 2M and the other has a concentration of 10M find the net direction of flow?

Solution:

Here given concentration and we know that concentration is directly proportional to osmotic pressure and osmotic pressure is proportional to number of solute particles and we know that net flow occurs from less solute particles to high solute particles. So net flow occurs from 2M to 10M. 


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Last Updated : 28 Nov, 2022
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