Question

When 100 g of ethylene polymerizes entirely to polyethene, the weight of polyethene formed as per the equation\[n(C{{H}_{2}}=C{{H}_{2}})\to -{{(C{{H}_{2}}-C{{H}_{2}})}_{n}}\] is:
A. $\dfrac{n}{2}g$
B. 100 g
C. $\dfrac{100}{n}g$
D. 100n g

Answer 1

Hint: The process of linkage of monomers to form the polymer is called polymerization. The weight of the polymer is going to depend on the number of monomeric units participated in the process of the polymerization.

Complete answer:
- In the question it is asked to calculate the weight of the polythene formed after the polymerization of the ethylene as per the chemical equation in the question.
- The given chemical reaction is as follows.
\[n(C{{H}_{2}}=C{{H}_{2}})\to -{{(C{{H}_{2}}-C{{H}_{2}})}_{n}}\]
- In the above chemical reaction ‘n’ moles of the ethylene is going to participate in the polymerization and forms a polythene as the product.
- The mass of ethylene is 100 g.
- By using the given weight of the ethylene, we can calculate the number of moles of the ethylene and it is as follows.
- The number of moles of the ethylene = $\dfrac{100g}{28}=3.5714mol$
- Therefore, the number of molecules of ethylene polymerized will be $3.5714\times {{N}_{A}}$ .
- Here ${{N}_{A}}$ = Avogadro number.
- Mass of the one ethylene molecule is 28 g/mol.
- Number of moles of ethylene units present in the polyethene will be $\dfrac{3.5714\times {{N}_{A}}}{{{N}_{A}}}=3.5714mol$ .
- Therefore, the mass of the 3.5714 mol of ethylene units = $3.5714\times 28=100g$ .
- So, the weight of the polythene is going to be formed as per the given chemical reaction is 100 g.

Therefore, the correct option is B.

Note:
We should know the molecular weight of the ethylene to calculate the total weight of the polymer which is going to form after polymerization of the ethylene. The molecular weight of the one ethylene molecule is 28 g.