--%>

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 : Finding Normality Can someone please

    Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N  (b) 19.8 N  (c) 39.6 N  (d) 98

  • Q : Problem on endothermic or exothermic At

    At low temperatures, mixtures of water and methane can form a hydrate (i.e. a solid containing trapped methane). Hydrates are potentially a very large source of underground trapped methane in the pole regions but are a nuisance when they form in pipelines and block th

  • Q : Modern periodic table and Mendeleevs

    Differentiate between the modern periodic table and Mendeleevs table?

  • Q : Preparation of ammonium sulphate Select

    Select the right answer of the question. Essential quantity of ammonium sulphate taken for preparation of 1 molar solution in 2 litres is: (a)132gm (b)264gm (c) 198gm (d) 212gm

  • Q : Coagulation what is the meaning of

    what is the meaning of fourth power of valency of an active ion?

  • Q : Problem based on molecular weight

    Select the right answer of the question. Molecular weight of urea is 60. A solution of urea containing 6g urea in one litre is : (a)1 molar (b)1.5 molar (c) 0.1 molar (d) 0.01 molar

  • Q : Molecular weight of substance The

    The boiling point of a solution of 0.11 gm of a substance in 15 gm of ether was found to be 0.1oC higher than that of the pure ether. The molecular weight of the substance will be (Kb = 2.16)       (a) 148 &nbs

  • Q : Problem on decinormal strength Can

    Can someone please help me in getting through this problem. How many grams of dibasic acid (having mol. wt. 200) must be present in 100ml  of its aqueous solution to provide decinormal strength: (i) 1g  (ii)2g  (iii) 10g  (iv) 20g<

  • Q : Explain the molecular mass with respect

    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

  • Q : What is Henry law constant and its

    1. The units of Henry Law constant are same as those of pressure, i.e. torr or h bar. 2. Different gases have dissimilar values of Henry law constant. The values of KH for some gases in water are given in tabl