Chapter 10.14, Problem 73AAP

To determine

The mole fraction of polyvlnyl chloride.

The mole fraction of polyvlnyl acetate.

Answer to Problem 73AAP

The mole fraction of polyvinyl chloride is $\underset{\_}{0.539}$.

The mole fraction of polyvinyl acetate is $\underset{\_}{0.461}$.

Explanation of Solution

Express the average molecular weight of the copolymer mer.

    $\begin{array}{c}{\text{MW}}_{\text{av}}\left(\text{mer}\right)={\text{f}}_{\text{PVC}}{\text{MW}}_{\text{PVC}}+{\text{f}}_{\text{PVA}}{\text{MW}}_{\text{PVA}}\\ ={\text{f}}_{\text{PVC}}{\text{MW}}_{\text{PVC}}+\left(\text{1}-{\text{f}}_{\text{PVC}}\right){\text{MW}}_{\text{PVA}}\end{array}$           (I)

Here, the mole fraction of polyvinyl chloride is ${\text{f}}_{\text{PVC}}$, the mole fraction of polyvinyl acetate is ${\text{f}}_{\text{PVA}}$.

Express the average molecular weight of the copolymer mer.

    ${\text{MW}}_{\text{av}}\left(\text{mer}\right)=\frac{{\text{MW}}_{\text{av}}\left(\text{polymer}\right)}{\text{DP}}$                   (II)

Here, the degree of the polymerization is DP.

Conclusion:

Substitute $11,000\text{g}/\text{mole}$ for ${\text{MW}}_{\text{av}}\left(\text{polymer}\right)$ and 150 mers for DP in Equation (II).

    $\begin{array}{c}{\text{MW}}_{\text{av}}\left(\text{mer}\right)=\frac{11,000\text{g}/\text{mole}}{\text{150}\text{mers}}\\ =73.33\text{g}/\text{mol}\cdot \text{mer}\end{array}$

The molecular weight of the PVC and PVA are as follows.

    $\begin{array}{l}{\text{MW}}_{\text{PVC}}=62.5\text{g}/\text{mol}/\text{mer}\\ {\text{MW}}_{\text{PVA}}=86.0\text{g}/\text{mol}/\text{mer}\end{array}$

Substitute $73.33\text{g}/\text{mol}\cdot \text{mer}$ for ${\text{MW}}_{\text{av}}\left(\text{mer}\right)$, $62.5\text{g}/\text{mol}/\text{mer}$ for ${\text{MW}}_{\text{PVC}}$, and $86.0\text{g}/\text{mol}/\text{mer}$ for ${\text{MW}}_{\text{PVA}}$ in Equation (I).

    $\begin{array}{l}\left(73.33\text{g}/\text{mol}\cdot \text{mer}\right)={\text{f}}_{\text{PVC}}\left(62.5\text{g}/\text{mol}/\text{mer}\right)+\left(\text{1}-{\text{f}}_{\text{PVC}}\right)\left(86.0\text{g}/\text{mol}/\text{mer}\right)\\ \left(73.33\text{g}/\text{mol}\cdot \text{mer}\right)=\left[\begin{array}{l}{\text{f}}_{\text{PVC}}\left(62.5\text{g}/\text{mol}/\text{mer}\right)\\ +\left(86.0\text{g}/\text{mol}/\text{mer}\right)-\left(86.0\text{g}/\text{mol}/\text{mer}\right){\text{f}}_{\text{PVC}}\end{array}\right]\\ {\text{f}}_{\text{PVC}}=0.539\end{array}$

Thus, the mole fraction of polyvinyl chloride is $\underset{\_}{0.539}$.

The mole fraction of PVA is obtained as follows,

    $\begin{array}{c}{\text{f}}_{\text{PVC}}=1-{\text{f}}_{\text{PVC}}\\ =1-0.539\\ =0.461\end{array}$

Thus, the mole fraction of polyvinyl acetate is $\underset{\_}{0.461}$.