Question

The molar mass distribution of a polymer is given as: 

Number of molecules	Molar mass (g/mol)
100	7500
50	5000

The resulting weight average molecular weight of the polymer is ................ g/mol. (Answer in integer)

Hint:

For polydisperse systems: - $M_n = \dfrac{\sum N_i M_i}{\sum N_i}$ (number average) - $M_w = \dfrac{\sum N_i M_i^2}{\sum N_i M_i}$ (weight average). Weight-average molecular weight gives more importance to heavier molecules.

Answer 1

Correct Answer: 6875

Step 1: Recall definition of weight-average molecular weight.
The formula is: \[ M_w = \frac{\sum N_i M_i^2}{\sum N_i M_i} \] where $N_i$ = number of molecules of type $i$, $M_i$ = molecular mass of that type.
Step 2: Substitute given data.
For $M_1 = 7500$, $N_1 = 100$: \[ N_1 M_1^2 = 100 \times (7500)^2 = 100 \times 56.25 \times 10^6 = 5.625 \times 10^9 \] \[ N_1 M_1 = 100 \times 7500 = 750000 \] For $M_2 = 5000$, $N_2 = 50$: \[ N_2 M_2^2 = 50 \times (5000)^2 = 50 \times 25 \times 10^6 = 1.25 \times 10^9 \] \[ N_2 M_2 = 50 \times 5000 = 250000 \]
Step 3: Compute numerator and denominator.
\[ \sum N_i M_i^2 = 5.625 \times 10^9 + 1.25 \times 10^9 = 6.875 \times 10^9 \] \[ \sum N_i M_i = 750000 + 250000 = 1,000,000 \]
Step 4: Calculate $M_w$.
\[ M_w = \frac{6.875 \times 10^9}{1.0 \times 10^6} = 6875 \ \text{g/mol} \] But wait — let's double check: - Numerator = $6.875 \times 10^9$ - Denominator = $1.0 \times 10^6$ Yes, result = 6875.

Step 5: Round to integer.
$M_w = 6875$ g/mol. Final Answer: \[ \boxed{6875 \ \text{g/mol}} \]