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Q.
If a polythene sample contains two monodisperse fractions in the ratio $2 : 3$ with degree of polymerization $100$ and $200$, respectively, then its weight average molecular weight will be :
Given,
Degree of polymerization of first polymer $=100$
Degree of polymerization of second polymer $=200$
$n _{1}=2$
$n _{2}=3$
Here the polymer is polythene which has a monomer ethylene with molecular mass $28$ ,
Now,
Degree of polymerization is given as:
$DP =\frac{\text { Molar mass of polymer }}{\text { molar mass of monomer }}$
Thus,
Molar mass of first polymer $M _{1}=100 \times 28=2800\, g / mol$
Molar mass of second polymer $M _{2}=200 \times 28=5600 \,g / mol$
Now, The weight average molecular weight is given as:
$\bar{M}_{w}=\frac{\sum n_{i} M_{i}^{2}}{\sum n_{i} M_{i}}$
Thus, putting all the values we get,
$\overline{ M }_{ w }=\frac{ n _{1} M _{1}^{2}+ n _{2} M _{2}^{2}}{ n _{1} M _{1}+ n _{2} M _{2}}$
$=\frac{2(2800)^{2}+3(5600)^{2}}{2 \times 2800+3 \times 5600}=4900\, g / mol$