A polymer sample contains three fractions: 120 chains of molar mass 15000 g/mol, 200 chains of 40000 g/mol and 80 chains of 90000 g/mol. Using $M_{n} = \sum N_{i} M_{i} / \sum N_{i}$ , find the number-average molar mass in kg/mol, correct to one decimal place.

Numerical answer type, answer in kg/mol, tolerance 0.05.

Answer: 42.5 kg/mol

M_{n} = \sum N_{i} M_{i} / \sum N_{i} = (120 \times 15000 + 200 \times 40000 + 80 \times 90000) /400

. The numerator is 17000000, so

M_{n} = 42500

g/mol

= 42.5

kg/mol.