--%>

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 : Problem on Redlich-Kwong equation i)

    i) Welcome to Beaver Gas Co.! Your first task is to calculate the annual gross sales of our superpure-grade nitrogen and oxygen gases. a) The total gross sales of N2 is 30,000 units. Take the volume of the cylinder to be

  • Q : Dipole moment of chloro-octane Describe

    Describe the dipole moment of chloro-octane in brief?

  • Q : Organic and inorganic chemistry Write

    Write down a short note on the differences between the organic and inorganic chemistry?

  • Q : Molar concentration of hydrogen 20 g of

    20 g of hydrogen is present in 5 litre of vessel. Determine he molar concentration of hydrogen: (a) 4  (b) 1 (c) 3 (d) 2 Choose the right answer from above.

  • Q : Problem on volumetric flow rate Methane

    Methane containing 4 mol% N2 is flowing through a pipeline at 105.1 kpa and 22 °C. To check this flow rate, N2 at the same temperature and pressure are introduced to the pipeline at the rate of 2.83 m3/min. At the end of the pipe (

  • Q : Numerical The volume of water to be

    The volume of water to be added to 100cm3 of 0.5 N N H2SO4 to get decinormal concentration is : (a) 400 cm3 (b) 500cm3 (c) 450cm3 (d)100cm3

  • Q : Organic and inorganic substances living

    living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?

  • Q : Molar and Volumetric flow rate problem

    Cyclohexane (C6H12) is produced by mixing Benzene and hydrogen. A process including a reactor, separator, and recycle stream is used to produce Cyclohexane. The fresh feed contains 260L/min C6H6 with 950 L/min of H2

  • Q : Production of alcoholic drinks give all

    give all physical aspects in the production of alcohol

  • Q : Strength of the Hydrochloric acid

    Provide solution of this question. 1.0 gm of pure calcium carbonate was found to need 50 ml of dilute HCL for complete reaction. The strength of the HCL solution is specified by : (a) 4 N (b) 2 N (c) 0.4 N (d) 0.2 N