- 7 5 0 Ro M2 6 5 M O Example 18 3 Determining The Size Of A Polymer By Light Scattering The Following Data For A Sample 1 (174.51 KiB) Viewed 35 times
7.5 0° RO)/m2 6.5 m O Example 18.3 Determining the size of a polymer by light scattering The following data for a sample of polystyrene in butanone were obtained at 20°C with plane-polarized light at 1= 546 nm. 26.0 36.9 66.4 90.0 113.6 19.7 18.8 17.1 16.0 14.4 In separate experiments, it was determined that K=6.42x10-5 mol m kg-2. From this information, calculate R. and Mw for the sample. Assume that the polymer is small enough that eqn 18.25 holds. Take cp = 311 kg m-3. Method As shown in the text, a plot of 1/R(O) against (sin2_0)/R(O) should be a straight line with slope 16tR/322 and y-intercept 1/Kc, Mw. Answer We construct a table of values of 1/R(0) and (sin?0)/R(O) and plot the data (Fig. 18.31). 010 26.0 36.9 66.4 90.0 113.6 {10-/R(O)}/m-2 5.08 5.32 5.85 6.25 6.94 {10²x (sin?0)/R(0)}/m2 2.57 5.33 17.5 31.3 48.6 The best straight line through the data has a slope of 0.388 and a y-intercept of 5.07 x 10-2. From these values and the value of K, we calculate R= 4.69x10-8m= 46.9 nm and Mw=987 kg mol-1. 100/(R(0)/m2) O р 5.5 О 4.5 0 10 20 30 40 50 {1000/R(0)/m2)} sin g Fig. 18.31 Plot of the data for Example 18.3.