If a polythene sample contains two monodisperse fractions in the ratio 2 : 3 with degree of polymerization 100 and 200, respectively, then its weight average molecular weight will be :

First, we need to find the weight average molecular weight of the polythene sample. Polythene is made from ethylene monomers, and the molecular weight of an ethylene monomer (C2H4) is calculated as follows: carbon has atomic weight 12 and hydrogen has atomic weight 1, so ethylene is (2 × 12) + (4 × 1) = 24 + 4 = 28 g/mol. Therefore, the molecular weight of the monomer, $$M_{\text{monomer}}$$, is 28 g/mol.

The degree of polymerization (DP) is the number of monomer units in a polymer chain. For the two fractions:

• Fraction 1 has DP = 100, so its molecular weight, $$M_1$$, is DP × $$M_{\text{monomer}}$$ = 100 × 28 = 2800 g/mol.
• Fraction 2 has DP = 200, so its molecular weight, $$M_2$$, is 200 × 28 = 5600 g/mol.

The sample contains these two monodisperse fractions in the ratio 2:3. In polymer chemistry, when such a ratio is given for monodisperse fractions without specification, it typically refers to the mole ratio (or number ratio) of the fractions. Therefore, we assume the mole ratio is 2:3, meaning the number of moles of fraction 1 to fraction 2 is 2:3.

Let the number of moles of fraction 1 be $$n_1 = 2k$$ and the number of moles of fraction 2 be $$n_2 = 3k$$, where $$k$$ is a constant. The total number of moles is $$n_{\text{total}} = n_1 + n_2 = 2k + 3k = 5k$$.

Now, we calculate the weight of each fraction:

• Weight of fraction 1, $$w_1 = n_1 \times M_1 = (2k) \times 2800 = 5600k$$ grams.
• Weight of fraction 2, $$w_2 = n_2 \times M_2 = (3k) \times 5600 = 16800k$$ grams.

The total weight of the sample is $$w_{\text{total}} = w_1 + w_2 = 5600k + 16800k = 22400k$$ grams.

The weight fraction of each fraction is the weight of that fraction divided by the total weight:

• Weight fraction of fraction 1, $$W_1 = \frac{w_1}{w_{\text{total}}} = \frac{5600k}{22400k} = \frac{5600}{22400}$$. Simplifying this fraction: divide numerator and denominator by 800 to get $$\frac{7}{28}$$, and again by 7 to get $$\frac{1}{4}$$. So, $$W_1 = \frac{1}{4}$$.
• Weight fraction of fraction 2, $$W_2 = \frac{w_2}{w_{\text{total}}} = \frac{16800k}{22400k} = \frac{16800}{22400}$$. Simplifying: divide numerator and denominator by 800 to get $$\frac{21}{28}$$, and again by 7 to get $$\frac{3}{4}$$. So, $$W_2 = \frac{3}{4}$$.

The weight average molecular weight, $$M_w$$, is calculated by multiplying the weight fraction of each fraction by its molecular weight and summing them up:

$$ M_w = (W_1 \times M_1) + (W_2 \times M_2) $$

Substitute the values:

$$ M_w = \left( \frac{1}{4} \times 2800 \right) + \left( \frac{3}{4} \times 5600 \right) $$

First, compute $$\frac{1}{4} \times 2800 = \frac{2800}{4} = 700$$.

Next, compute $$\frac{3}{4} \times 5600 = \frac{3 \times 5600}{4} = \frac{16800}{4} = 4200$$.

Now, add them together: $$M_w = 700 + 4200 = 4900$$ g/mol.

Hence, the weight average molecular weight is 4900 g/mol. Comparing with the options, this corresponds to option A.

So, the correct answer is Option A.