Discussion
If the mixture contains equal number of molecules of A, B and C, the weight average molecular weight, \(\overline{M_{w}}\) will be
Answer: C
The weight average molecular weight is \(2.33 \times 10^{4}\) g mol-1.
With equal numbers of molecules, take one of each:
\[ \begin{aligned} \overline{M_{w}} &= \frac{\sum N_{i}M_{i}^{2}}{\sum N_{i}M_{i}} = \frac{(1 + 4 + 9) \times 10^{8}}{(1 + 2 + 3) \times 10^{4}} \\ &= \frac{14 \times 10^{8}}{6 \times 10^{4}} = 2.33 \times 10^{4} \end{aligned} \]
The number average for this mixture is simply 20000, and \(\overline{M_{w}}\) is higher because it weights the heavier molecules more. 1.93 and 2.13 fall too close to the number average, and 2.53 overweights the heaviest component.
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