If a polythene sample contains two monodisper | vedclass

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Question

(A) Given,Degree of polymerization of first polymer $= 100$Degree of polymerization of second polymer $= 200$Number of moles ratio $n_1 : n_2 = 2 : 3$

Vedclass · Accepted Answer

$4900$

Vedclass · Answer

(A) Given,Degree of polymerization of first polymer $= 100$Degree of polymerization of second polymer $= 200$Number of moles ratio $n_1 : n_2 = 2 : 3$The monomer of polythene is ethylene $(C_2H_4)$ with molar mass $M_0 = 28 \ g/mol$.Molar mass of first polymer fraction: $M_1 = 100 	imes 28 = 2800 \ g/mol$.Molar mass of second polymer fraction: $M_2 = 200 	imes 28 = 5600 \ g/mol$.The weight average molecular weight is given by the formula:$\overline{M}_w = \frac{\sum n_i M_i^2}{\sum n_i M_i} = \frac{n_1 M_1^2 + n_2 M_2^2}{n_1 M_1 + n_2 M_2}$Substituting the values:$\overline{M}_w = \frac{2 	imes (2800)^2 + 3 	imes (5600)^2}{2 	imes 2800 + 3 	imes 5600}$$\overline{M}_w = \frac{2 	imes 7840000 + 3 	imes 31360000}{5600 + 16800}$$\overline{M}_w = \frac{15680000 + 94080000}{22400}$$\overline{M}_w = \frac{109760000}{22400} = 4900 \ g/mol$.

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

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