Question

A polymer sample is a mixture of 1 mole of each of two macromolecules with molar masses 20000 and 50000 respectively. The ${{M}_{n}}$ of the sample is
(A) 20000
(B) 50000
(C) 35000
(D) 70000

Answer 1

Hint: When the polymer is made up of two or more macromolecules, the average mass of the resulting polymer is calculated as ${{M}_{n}}$. As we calculate the average of any quantity, this too is calculated the same.

Complete step by step solution:
Let us define the average and ${{M}_{n}}$ for polymers before solving the given problem;
Average- Average of any measurable quantity is the ratio of the total amount of the quantities to the number of quantities. For example, let us consider we have three samples having weights as 1 gm, 2 gm and 3 gm each. Thus,
\[Average=\dfrac{1+2+3}{3}=2gm\]
Similarly, ${{M}_{n}}$ is known as the number average molecular weight of a polymer. Generally, ${{M}_{n}}$ can be predicted by the polymerisation mechanisms and is measured by the method that would determine the number of molecules in a sample of given weight. In short, ${{M}_{n}}$ is just the total weight of all the polymer molecules in the sample, divided by the number of the total number of polymer molecules in the same.
Thus, by using the prior knowledge of ${{M}_{n}}$ we can solve as,
Given that,
Moles of macromolecules = 1 each
Molar mass of a macromolecule = 20000 units
Molar mass of another macromolecule = 50000 units
Thus, ${{M}_{n}}$ can be given as,
$\begin{align}
& {{M}_{n}}=\dfrac{\left( 1\times 20000 \right)+\left( 1\times 50000 \right)}{1+1} \\
& {{M}_{n}}=\dfrac{70000}{2}=35000units \\
\end{align}$
Thus, the ${{M}_{n}}$ of a sample consisting of two macromolecules is 35000 units.

Therefore, option (C) is correct.

Note: Note that the calculation related to ${{M}_{n}}$ would be based on the given data. If a degree of polymerisation is given, the calculation scenario will change.