If the number-average degree of polymerization for polystyrene obtained by the bulk polymerization of styrene at 60 (C) is 1000, what would be the number-average degree of polymerization if the polymerization were conducted in a 10% solution in toluene (900 g of toluene per 100 g of styrene) under otherwise identical conditions? The molecular weights of styrene and toluene are 104.12 and 92.15, respectively. State any assumptions that are needed.

Respuesta :

Explanation:

Formula to calculate degree of polymerization is as follows.

        Degree of polymerization (DP) = [tex]\frac{\text{Number average molecular weight}}{\text{Average unit molecular weight}}[/tex]

         DP = [tex]\frac{\bar{M_{n}}}{m}[/tex]

        1000 = [tex]\frac{\bar{M_{n}}}{104.12}[/tex]

       [tex]\bar{M_{n}}[/tex] = 104150

Formula to calculate average repeat molecular weight of co-polymer is as follows.

              [tex]\bar{m} = \sum{f_{j}m_{j}}[/tex]

where,    [tex]f_{j}[/tex] = mole fraction of each component

Moles of styrene = [tex]\frac{mass}{\text{molar mass}}[/tex]              

                            = [tex]\frac{100 g}{104.12 g/mol}[/tex]

                            = 0.96 mol

Moles of toulene = [tex]\frac{900 g}{92.15 g/mol}[/tex]

                            = 9.77 mol

Therefore, mole fraction will be calculated as follows.

           [tex]f_{1} = \frac{0.96}{0.96 + 9.77}[/tex]

                   = 0.0895

           [tex]f_{2} = \frac{9.77}{0.96 + 9.77}[/tex]

                   = 0.910

    [tex]\bar{m} = 0.0895 \times 104.12 g/mol + 0.910 \times 92.15 g/mol[/tex]

               = 93.22 g/mol

Hence, DP = [tex]\frac{\bar{M_{n}}}{\bar{m}}[/tex]

                  = [tex]\frac{104150}{93.22 g/mol}[/tex]

                  = 1117

Thus, we can conclude that the number-average degree of polymerization is 1117.