Question

A 5-g sample of a polyester having one carboxyl group per molecule is to be titrated by sodium hydroxide solutions to determine its number average molecular weight. How much 0.01 molar sloution is required if the polymer has molecular weight of approximately 1000 g/mol? 10,000g/mol? 100,000 g/mol? Discuss the practicality of the experiment, with special reference to the upper molecular weight limit that can be analyzed.

Answer 1

   no of moles = given mass / molar mass = 5g/ 1000g/mol = 0.005 mol

1) number average molecular weight = no of moles * molecular weight / molecular weight

   = 0.005  * 1000/ 1000 = 0.005  mole

ii) As

M₁V1 (polymer)=M₂V₂(NaoH) = 0.005 * 1000 ml = 0.01 * V₂

V₂ = 500 ml (for 1000 g/mol )

2)    no of moles = given mass / molar mass = 5g/ 10000 g/mol = 0.0005 mol

M₁V1 (polymer)=M₂V₂(NaoH) = 0.0005 * 1000 ml = 0.01 * V₂

V₂ = 50 ml (for 10000 g/mol )

3)

no of moles = given mass / molar mass = 5g/ 100000 g/mol = 0.00005 mol

M₁V1 (polymer)=M₂V₂(NaoH) = 0.00005 * 1000 ml = 0.01 * V₂

V₂ = 5 ml (for 100000 g/mol )