Question

Titration was used to determine the molar mass of two linear monodisperse polymers A and B. Both polymers have the same repeat unit and contain acid end-groups. - 10 g of polymer A titrated with 5 mL of 0.1 M alkali solution. - 5 g of polymer B titrated with 5 mL of 0.1 M alkali solution. - All alkali reacted with acid end-groups. Find the ratio of the molar mass of A to that of B. (Round off to one decimal place)

Hint:

For end-group analysis, molar mass is obtained by dividing polymer mass by moles of titrant consumed.

Answer 1

Correct Answer: 2

Step 1: Moles of alkali used.
\[ n_{NaOH} = M \times V = 0.1 \times 0.005 = 0.0005 \; \text{mol} \] This equals the number of polymer molecules (since 1 end-group per chain).

Step 2: Number of moles of polymer chains.
- For Polymer A: \(n_A = 0.0005 \; \text{mol}\). - For Polymer B: \(n_B = 0.0005 \; \text{mol}\).

Step 3: Molar mass of each polymer.
\[ M_A = \frac{\text{mass}}{\text{moles}} = \frac{10}{0.0005} = 20000 \; g/mol \] \[ M_B = \frac{5}{0.0005} = 10000 \; g/mol \]

Step 4: Ratio.
\[ \frac{M_A}{M_B} = \frac{20000}{10000} = 2.0 \] Final Answer: \[ \boxed{2.0} \]