Discussion
For a polydisperse polymer, the viscosity-average molecular weight \((\bar{M}_{v})\) becomes equal to the weight-average molecular weight \((\bar{M}_{w})\), when the Mark-Houwink constant a is ___.
Answer: D
\(\bar{M}_{v} = \bar{M}_{w}\) when a = 1.0.
The viscosity-average molecular weight is defined through the Mark-Houwink exponent a:
\[ \bar{M}_v^{\,a} = \frac{\sum N_i M_i^{1+a}}{\sum N_i M_i} \]
With a = 1 this becomes \(\sum N_{i}M_{i}^{2}/\sum N_{i}M_{i}\), which is exactly the weight average. For the usual values of a, between 0.5 (theta solvent) and 0.8 (good solvent), \(\bar{M}_{v}\) lies between the number and weight averages, closer to the weight average, so 0.5 and 0.7 do not make them equal; 0.1 is an unrealistic exponent for flexible coils.
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