1.)  Easiest way to answer this question is to use a spreadsheet to create a series for M_i from 1000 to 100,000 stepping by 100 or smaller.  Create another column that is equal to the weight w_i, which is constant. 

Calculate N_i by dividing w_i/M_i.  Use the equations and below and the spreadsheets to calculate Mn and Mw  

Mn = S N_iM_i/S N_I = 21,470  

 

Mw = S w_iM_i / S w_i = 50,500  

Alternatively, you can transform the summations to integrals and solve.   For instance in integral form, M_w = , where K is the height (y-value) of the curve shown in the question    

 

2.) D_n = S N_iD_i/S D_I = 2.54 cm; where D_i and N_i are given in the table  

D_w = S w_iD_i / S w_i = 3.38 cm; Here, w_i is not the same as for MW analysis.  w_i is related to the volume of each sphere.  You needed to go back to the original formula, calculate the weight of each sphere and plug into this equation.  The weight of each sphere is 4/3 p R³ * density.   

 

3.) p = r g h , calculate p for each height in the correct units (mol/g).  Plot vs. concentration (in correct units) to get a straight line.  Intercept give 1/Mn à Mn = 189,000; Slope gives A₂ = .00057 mol cm³ /g²  

4.) Plot 1/C (t/t_o –1) vs. concentration to get Huggins constant from slope and intrinsic viscosity (I.V) from Intercept.  Similar plot of 1/C ln(t/t₀) also gives a straight line with same intercept, which is I.V.; k_h = .27, I.V. = .876 dl/g.From a number of sources, K = 9.2 * 10^-5 dl/g, a =0.72.  Plug this K, a, and I.V. into MHS equation to get M_v = 332,000   I also accepted K = 12.0 * 10^-5 dl/g and a = 0.71 from the polymer handbook, which gives M_v = 276,000  

 

5.) From problem 1, for a given w_I, N_i = w_i/M_i , thus for a GPC curve that gives you a plot of w_i vs. M_i (if calibrated correctly!) you can use the analysis as given in Question 1 to get the M_n and M_w of the sample.  

 

6.) This question is answered by calculating sin²q/2 + bc for each data point given in the table, plotting Kc/R_q vs sin²q/2 + bc, extrapolating each line to q=0 or c=0 to get points that define the lines from which M_w is obtained from the intercept, A₂ is obtained from the slope of the q=0 line and the R_g of the polymer is obtained from the slope of the c=0 line.  An example of this procedure is given in your book, Example 8.3 on Page 309.  I expected you to provide M_w (741,000) and A₂ (4.5 * 10^-4 cm³ mol g^-2). With the assumption of using a 546 nm laser, you could estiamte R_g (48 nm). I did not take points off if you did not calculate R_g, however, points were taken off if you did not mention that you could determine R_g with a little more information.