JEE Advanced 2022 Paper 1 · Chemistry · Biomolecules and Polymers · medium

A sample of a synthetic polymer contains two fractions of different chain lengths: 60% by mass of chains of molar mass 20000 g mol^-1 and 40% by mass…

Numerical answer

A sample of a synthetic polymer contains two fractions of different chain lengths:

60\%

by mass of chains of molar mass

20000\ \mathrm{g\,mol^{-1}}

and

40\%

by mass of chains of molar mass

50000\ \mathrm{g\,mol^{-1}}

. The polydispersity index

\overline{M_w}/\overline{M_n}

of the sample is (give the answer correct to two decimal places)

Show answer and solution

Answer: 1.22

The mass fractions

w_i

are given, not mole fractions. The weight-average molar mass is

\overline{M_w}=\sum w_iM_i=0.60(20000)+0.40(50000)=32000\ \mathrm{g\,mol^{-1}}

. For the number-average, per gram of sample the moles of chains of each type are

w_i/M_i

, so

\overline{M_n}=\dfrac{\sum m_i}{\sum n_i}=\dfrac{1}{\sum w_i/M_i}=\dfrac{1}{\dfrac{0.60}{20000}+\dfrac{0.40}{50000}}=\dfrac{1}{3.0\times10^{-5}+0.8\times10^{-5}}=26316\ \mathrm{g\,mol^{-1}}

. Polydispersity index

=32000/26316=1.216\approx1.22

. Using the mass fractions as mole fractions would give

\overline{M_n}=32000

and an index of exactly

1

, which is the main trap. A value above

1

shows the chains are not uniform. Answer: 1.22.

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