--%>

Explain the molecular mass with respect to polymers.

During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is taken. There are two kinds of average molecular masses of polymers.

    
1. Number-average molecular mass  2454_polymers1.png 
    
2. Mass-average molecular mass  2192_Polymers2.png 

The two types of molecular masses are defined and calculated as follows:
    
1. Number-average molecular mass

When the total mass of all the molecules of a sample is divided by the total number of molecules, the result obtained is called the number-average molecular mass. For example, suppose in a particular sample

N1 molecules have molecular mass M1 each.

N2 molecules have molecular mass M2 each.

N3 molecules have molecular mass M3 each and so on. Then, we have

Total mass of all the N1 molecules = N1M1.

Total mass of all the N2 molecules = N2M2.

Total mass of all the N3 molecules = N3M3 and so on.

 Total mass of all the molecules = N1M1 + N2M2 + N3M3 + .....

= ΣNiMi

Total number of all the molecules = N1 + N2 + N3 + ....

= ΣNi

Hence the number-average molecular mass will be given by
732_Polymers3.png 


1827_polymers1.png is generally determined by osmotic pressure measurement.
    
2. Mass-Average molecular mass

When the total mass of groups of molecules having different molecular masses are multiplied with their respective molecular masses, the products are added and the sum is divided by the total mass of all the molecules, the result obtained is called the mass-average molecular mass. Supposing, as before that N1N2N3, etc, molecules have molecular mass M1M2M3 etc. correspondingly.

Total mass of N1 molecules = N1M1.

Total mass of N2 molecules = N2M2.

Total mass of N3 molecules = N3M3 and so on.

The products with their respective molecular masses will be (N1M1 × M1)(N2M2 × M2)(N3M3 × M3), etc. i.e. N1M12N2M22N3M32, etc.

Sum of the products = N1M12 + N2M22 + N3M32 + ......

= ΣNiMi2

Hence the mass-average molecular mass is given by
879_Polymers4.png 


2192_Polymers2.png is generally determined by technique like ultra centrifugation of sedimentation.

 

 

 

 

 

 

 

   Related Questions in Chemistry

  • Q : What are various structure based

    This classification of polymers is based upon how the monomeric units are linked together. Based on their structure, the polymers are classified as: 1. Linear polymers: these are the polymers in which monomeric units are linked together to form long straight c

  • Q : Sedimentation and Velocity The first

    The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A

  • Q : Relationship between free energy and

    The free energy of a gas depends on the pressure that confines the gas. The standard free energies of formation, like those allow predictions to be made of the possibility of a reaction at 25°C for each reagent at 

  • Q : What do you mean by the term enzymes

    What do you mean by the term enzymes? Briefly illustrate it.

  • Q : Distribution law Help me to go through

    Help me to go through this problem. The distribution law is applied for the distribution of basic acid between : (a) Water and ethyl alcohol (b) Water and amyl alcohol (c) Water and sulphuric acid (d) Water and liquor ammonia

  • Q : Atmospheric pressure Give me answer of

    Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules

  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

  • Q : Neutralization of sodium hydroxide How

    How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g  (ii) 6 g (iii) 40 g  (iv) 60 g

  • Q : Application of colligative properties

    Choose the right answer from following. Colligative properties are used for the determination of: (a) Molar Mass (b) Equivalent weight (c) Arrangement of molecules (d) Melting point and boiling point (d) Both (a) and (b)  

  • Q : Problem based on mole concept Choose

    Choose the right answer from following. An aqueous solution of glucose is 10% in strength. The volume in which mole of it is dissolved will be : (a) 18 litre (b) 9 litre (c) 0.9 litre (d) 1.8 litre