Polymers and rubbers

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Question

A polymer contains 10 molecules with molecular mass 10,000 and 10 molecules with molecular mass 1,00,000 calculate the polydispersity index of the polymer

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Solution

$P . D . I = \frac{{\bar{M}}_{w}}{{\bar{M}}_{m}}$
${\bar{M}}_{w} = \frac{\sum_{i = 1}^{α} N_{i} M_{i}^{2}}{\sum_{i = 1}^{α} N_{i} M_{i}} = \frac{10 \times {(10^{4})}^{2} + 10 \times {(10^{5})}^{2}}{10 \times (10^{4}) + 10 \times (10^{5})} = 91, 818.18$
${\bar{M}}_{m} = \frac{\sum_{i = 1}^{α} N_{i} M_{i}}{\sum_{i = 1}^{α} N_{i}} = \frac{10 \times 10^{4} + 10 \times 10^{5}}{10 + 10} = 55, 000$
$∴ P D I = \frac{91, 818, 18}{55, 000} = 1.669 ≅ 1.67$

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Polymers and rubbers

Remember concepts with our Masterclasses.

A polymer contains 10 molecules with molecular mass 10,000 and 10 molecules with molecular mass 1,00,000 calculate the polydispersity index of the polymer

P.D.I=M¯wM¯mM¯w=∑i=1αNi Mi2∑i=1αNi Mi=10×1042+10×105210×104+10×105=91,818.18M¯m=∑i=1αNi Mi∑i=1αN i=10×104+10×10510+10=55,000∴ PDI=91,818,1855,000=1.669≅1.67

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