In [3]:

```
from __future__ import division
#given
#degree of polymerization of styrene
DOP=10000
#formula of styrene= C8H8
#molecular weight of styrene monomer
Mm=12*8+1*8
#molecular weight of polymer
Mp=DOP*Mm
print "the molecular weight of styrene polymer is %d"%(Mp)
```

In [7]:

```
#given
#degree of polymerization of teflon
DOP=100000
#chemical formula of teflon is C2F4
#molecular weight of monomer teflon
Mm=2*12+4*19
#molecular weight of teflon polymer
Mp=DOP*Mm
#molecular weight of polythene monomer i.ee C2H4
MmP=2*12+4*1
#molecular weight of polythene polymer
MpP=DOP*MmP
print "molecular weight of PTFE anf Polythene are %d and %d"%(Mm,MpP)
#ratio of molecular weight of PTFE and Polythene
R=Mp/MpP
print "the ration of molecular weight of PTFE and Polythene having same DOP is %0.3f"%(R)
```

In [5]:

```
#given
#the molecular weight of polyisoprene monomer
Mm=68 #gm
#after vulcanisation with sulphur, it is observedthat the 2 molecules of isoprene monimer require 2 molecules of sulphur
#hence for full cross linking ,(68*2) gm of isoprene requires (32*2)gm of sulphur. therefore 68kg of isoprene requires
M=32*2*68/68/2 #kg of sulphur
print "the weight of sulphur required for cross link polymerization of polyisoprene is %d of sulphur"%(M)
print "the fully cross linked product will be EBONITE"
```

In [6]:

```
from __future__ import division
#given
#molecular weight of butdiene,isoprene,sulhur and carbon black are
MB=4*12+6*1
MI=68
MS=32
MC=12
#percentages of diiferent constituents in rubber
PB=27/MB
PI=51/MI
PS=16/MS
PC=6/MC
#percentage of coss linking
percent=PS/(PI+PB)*100
print "the percentage of cross linking is %d"%(percent)
```