A sample consists of two monodisperse fractions: $30%$ by mass has $M = 20000$ and $70%$ by mass has $M = 80000 g mo l^{- 1}$ . Using $M_{n} = \sum N_{i} M_{i} / \sum N_{i}$ , $M_{w} = \sum N_{i} M_{i}^{2} / \sum N_{i} M_{i}$ and PDI $= M_{w} / M_{n}$ , the PDI is

Answer: B

Take 100 g: moles of fractions

= 30/20000 = 1.50 \times 1 0^{- 3}

and

70/80000 = 8.75 \times 1 0^{- 4}

M_{n} = 100/ (2.375 \times 1 0^{- 3}) = 42105

M_{w} = 0.30 (20000) + 0.70 (80000) = 62000

(mass-weighted). PDI

= 62000/42105 = 1.47