Question

Titration was used to determine the molar mass of two linear monodisperse polymers A and B. Both the polymers possess the same repeat unit and contain acid end-groups. First, 10 g of polymer A was titrated with 5 mL of a 0.1 M alkali solution. In a separate experiment, 5 g of polymer B was titrated with 5 mL of a 0.1 M alkali solution. All of the alkali solution reacted with the acid end-groups present in both polymer A and polymer B.\ The ratio of the molar mass of A to the molar mass of B is ............ {(Round off to one decimal place)}

Hint:

For molar mass determination from titration, use: \( {Molar Mass} = \frac{{Mass}}{{Moles of reacting end-groups}} \).

Answer 1

Given:
- Volume of alkali used for both = 5 mL = 0.005 L
- Molarity = 0.1 M, so moles of alkali = \(0.005 \times 0.1 = 5 \times 10^{-4}\) mol
- Moles of polymer = moles of alkali (1:1 reaction)
Let molar masses be \(M_A\) and \(M_B\). \[ M_A = \frac{10}{5 \times 10^{-4}} = 20000 {and} M_B = \frac{5}{5 \times 10^{-4}} = 10000 \] So the ratio \( \frac{M_A}{M_B} = \frac{20000}{10000} = 2.0 \)