A sample contains 2 mol of chains of molar mass $15000 g mo l^{- 1}$ and 1 mol of chains of molar mass $45000 g mo l^{- 1}$ . Use $M_{w} = \sum N_{i} M_{i}^{2} / \sum N_{i} M_{i}$ . The weight-average molar mass ( $g mo l^{- 1}$ ) is

Answer: D

M_{w} = (2 \times 1500 0^{2} + 1 \times 4500 0^{2}) / (2 \times 15000 + 1 \times 45000) = (2475000000) / (75000) = 33000

. The number-average

25000

and the simple mean

30000

are common errors.