Chapter 23, Problem 1QAP

(a)

Interpretation Introduction

Interpretation:

A portion of the polymer made from tetrachloroethylene needs to be drawn.

Concept Introduction:

A polymer is a long chain consists of large number of monomer units. In a polymer, the monomers are linked to each other in a continuous or repetitive manner. These monomer units are linked to each other either through the formation of peptide linkage,glycosidic linkage or by removal of any moiety such as a water molecule.Polyvinyl chloride, Bakelite and polystyrene are some of the examples of the polymers.

Answer to Problem 1QAP

A portion of polymer made from tetrachloroethylene is as follows:

  

Explanation of Solution

Tetrachloroethylene polymer consists of number of monomer unit of 1,1,2,2-tetrachloroethylene.

The structure of tetrachloroethylene or monomer unit of the polymer is as follows:

  

The polymer is formed after breaking of C-C double bonds, the two terminal carbon atoms of 1 tetrachloroethylene unit form bond with 2 terminal carbons of two other tetrachloroethyleneunits. To form a polymer, n numbers of such tetrachloroethyleneunits combined in the same way.

A portion of this polymer is shown as follows:

  

(b)

Interpretation Introduction

Interpretation:

The molar mass of the polymer needs to be calculated if it contains $3.2\times {10}^3$ tetrachloroethylene molecules.

Concept Introduction:

A polymer is a long chain consists of large number of monomer units. In a polymer, the monomers are linked to each other in a continuous or repetitive manner. These monomer units are linked to each other either through the formation of peptide linkage, glycosidic linkage or by removal of any moiety such as a water molecule. Polyvinyl chloride, Bakelite and polystyrene are some of the examples of the polymers.

Tetrachloroethylene polymer consists of number of monomer unit of 1,1,2,2-tetrachloroethylene.

Answer to Problem 1QAP

The molar mass of complete polymer is $\text{ 5}\text{.38 }\times {\text{ 10}}^{\text{5}}\text{g /mol}$ .

Explanation of Solution

Molar mass of 1 monomer unit of 1,1,2,2-tetrachloroethylene is 168 g /mol. This is mass of 1 molecule of the 1,1,2,2-tetrachloroethylene.

If the polymer contains $\text{3}\text{.2 }\times {\text{ 10}}^{\text{3}}$ monomer unit of 1,1,2,2-tetrachloroethylene. The molar mass of the polymer can be calculated as follows:

  $M_{\text{polymer}}\text{= }\left(\text{3}\text{.2 }\times {\text{ 10}}^{\text{3}}\right)\times \text{168 g/mol = 5}\text{.38}\times {\text{10}}^{\text{5}}\text{ g /mol}$

Thus, molar mass of complete polymer is $\text{5}\text{.38 }\times {\text{ 10}}^{\text{5}}\text{g /mol}$ .

(c)

Interpretation Introduction

Interpretation:

The mass percent of C and Cl in the polymer needs to be determined.

Concept Introduction:

Mass percent of an atom present in the sample can be determined by dividing mass of atom present in the monomer to the overall mass of the monomer unit and multiplying the overall result with 100%.

For example, the mass percent of x g of an atom present in the y g of monomer unit can be determined as:

  $M\text{ \% = }\left(\frac{x\text{ g}}{y\text{ g}}\right)\times \text{100\%}$

Answer to Problem 1QAP

Mass percent of C in polymer is 14.28 %.

Mass percent of Cl in polymer is 83.33 %.

Explanation of Solution

The molar mass of the 1 tetrachloroethylene unit is 168 g/mol.

There are 2 C atoms in a monomer unit of tetrachloroethylene.

Now, molar mass of C atom in a monomer will be:

  $M_{\text{C}}\text{= 2}\times \text{12 g/mol=24 g/mol}$

  $\%m_{\text{C}}=\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\times 100\%$

Putting the values,

  $\%m_{\text{C}}=\frac{\text{24 g/mol}}{\text{168 g/mol}}\times \text{100\% = 14}\text{.28 \%}$

Thus, mass percent of C in polymer is 14.28 %.

The number of Cl atoms in the monomer unit of tetrachloroethylene is 4.

Molar mass of Cl in a monomer will be:

  $M_{\text{Cl}}\text{= 4 }\times \text{ 35}\text{.5 = 140 g /mol}$

Mass percent of Cl can be calculated as follows:

  $\%m_{\text{Cl}}=\frac{M_{\text{Cl}}}{M_{\text{monomer unit}}}\times 100\%$

Putting the values,

  $\%m_{\text{Cl}}\text{= }\frac{\text{140 g/mol}}{\text{168 g/mol}}\times \text{100\% = 83}\text{.33 \%}$

Thus, mass percent of Cl in the polymer is 83.33 %