Ncert Class 12th Solution part-2

Chemistry Chapter -Solution Part-2

How do you find the molar mass of Colligative properties?
How do you find the molar mass of a solute?
How do you determine Colligative properties?
How do you find the molecular weight of an unknown compound?

Topics to be covered in this Post 

Molality
Molarity
Normality
Mole fraction
Colligative properties 

Molality:-  It is defined as the total moles of solute dissolved in the weight of the solvent .

                Molality =Number of moles of solute/weight of solvent in (k.g.)
                                    It is denoted by "m" 
                                 Its unit is moles/kg.

Molarity:- It is defined as total number of moles of solute dissolved in the volume of the solution.

              Molarity =Number of moles of solute/volume of solution(ltr)

                    It is denoted by "M"
                      its unit is moles/ltr

Normality = It is defined as the number of gram equivalent of solute in a volume of the solution.

Normality =No.of gram equivalent of the solute/Volume of the solution

NO.of Gram equivalent of the solute = Given the weight of the solute/equivalent weight of the solute.

Equivalent weight =Molecular mass of the solute /valency,acidity or basicity

Mole fraction:- It is defined as the ratio of moles of a particular component with the total number of components of the solution.

Mole fraction of Component A =Xa/Xa+Xb+Xc

Colligative properties:- 

The properties of a solution which depends on the number of a solute particle but not depend on the nature of the solute particle.These types of properties known as colligative properties.

(1) Relative lowering in vapor pressure 

(2) Elevation in Boiling point
(3)Depression in freezing point.

 Relative lowering in Vapor pressure:- 

In a solution solid non-volatile solute present over the surface of the solvent, the vapor pressure of the solvent is less than the pure state.

Expression for  Relative lowering in vapor pressure :-
P/Pa=Xa
1-P/Pa=1-Xa
Pa-p/Pa=Xb
OR
Pa>P
        ΔP=Pa-P                                         (P=Pa=PaXa)
ΔP=Pa-PaXa                     
ΔP=Pa(1-Xa)
ΔP/Pa=1-Xa                                          ΔP=Pa-P)
Pa-P/Pa=Xb                                          (Xa+Xb=1)

So the relative lowering of the vapor pressure of the solution is directly depends upon the mole of solute or non-volatile component of the solution.

How to Calculate the Molecular mass of the solute By relative lowering of Vapor pressure ?

Pa-P/Pa=Xb
Pa-P/Pa=Nb/Na                    (Na is the moles of solvent )
                                                     (Nb is the moles of Solute )
ΔP/Pa=Wb/Mb*Ma/Wa                                  Wb=Weight of solute in gram
                                                                                   Wa= weight of the solvent in gram
Mb=Wb.Ma*Pa/Wa*ΔP

Elevation in Boiling point:-  

The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to 1 atm.

In a solution of non-volatile solute (B) in liquid (A) the vapor pressure of the solute ion is always less than the vapour pressure of the pure solvent. So, the boiling point of the solution is more than a normal boiling point of the solution.

ΔTb=Kb.m                 (m= is the molality of the solution .Kb is the elevation constant )
How to determine the molecular mass of the solute by Elevation in the boiling point method?
ΔTb=Kb.m
ΔTb=Kb.Nb*1000/Wa
ΔTb=KbWb*1000/Mb.Wa
Mb=Kb.Wa*1000/ΔTb.Wa

Depression in freezing point:-

  The freezing point is the temperature at which the vapor pressure of the solid phase is equal to vapour pressure of the substance.

ΔTf=Tf'-Tf

ΔTf=Kf.m                                             (m is the molality
                                                                 Kf is the molal freezing point depression constant )

How todetermine the molecular mass of the solute by Depression in freezing point method?

   ΔTf=Kf.Nb/Wa*1000
ΔTf=Kf*Wb*1000/Mb.Wa                                           (Wa is the weight of the solvent A,
                                                                                                Wb weight of the solute in gram B
                                                                                                Mb is the mass of solute B )
Mb=Kf*Wb*1000/ΔTf*Wa

Osmotic pressure:- 

The extra pressure which is applied to a solute ion to prevent the movement of solvent particles from low concentrated solution to higher concentrated solution through a semi-permeable membrane

It is denoted by π
              π=CRT                     (C is the concentration od solution )
π=Nb.RT/V
π=Wb*RT/Mb.V
Mb=Wb*RT/π.V

Isotonic solution:- Two or more solutions that have similar concentration as well as similar osmotic pressure.

e.g. 1M HCL and 1M NaCl.

Hypotonic Solution:- The solution which has lower osmotic pressure and concentration than other solution.

e.g. 1 m Hcl solution behaves as a hypotonic solution with 2M NaCl.

Hypertonic solution:-The solution which has a higher concentration and higher osmotic pressure than other solution.

e.g. 2M Hcl behaves as a hypertonic solution with 1 M Hcl.

Reverse Osmosis:- The pressure of movement of solvent particles from the higher concentrated solution to lower concentrated solution through a semi-permeable membrane, by applying extra pressure on a

higher concentrated solution.

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