A polymer sample contains 400 chains of molar mass 2.0x10^4, 400 chains of 5.0x10^4 and 200 chains of 1.0x10^5 g mol^-1. Use M_n=sum N_iM_i/sum N_i, M_w=sum N_iM_i^2/sum N_iM_i and PDI =M_w/M_n. Find the PDI, to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3pl165

A polymer sample contains 400 chains of molar mass $2.0 \times 1 0^{4}$ , 400 chains of $5.0 \times 1 0^{4}$ and 200 chains of $1.0 \times 1 0^{5} g mo l^{- 1}$ . Use $M_{n} = \sum N_{i} M_{i} / \sum N_{i}$ , $M_{w} = \sum N_{i} M_{i}^{2} / \sum N_{i} M_{i}$ and PDI $= M_{w} / M_{n}$ . Find the PDI, to two decimal places.

Numerical answer type, tolerance 0.01.

\sum N_{i} M_{i} = 400 (20000) + 400 (50000) + 200 (100000) = 4.8 \times 1 0^{7}

and

\sum N_{i} = 1000

, so

M_{n} = 48000

\sum N_{i} M_{i}^{2} = 1.6 \times 1 0^{11} + 1.0 \times 1 0^{12} + 2.0 \times 1 0^{12} = 3.16 \times 1 0^{12}

, so

M_{w} = 3.16 \times 1 0^{12} /4.8 \times 1 0^{7} = 65833

. PDI

= 65833/48000 = 1.37