Welcome to the forum. It is a rule that you must show an attempt before we can help you. However, this problem is constructed in an unusual way, so it might take a little bit more thought than a typical titration problem. Maybe you can give an outline of how you think you should proceed, or some more general thoughts.
Thanks!
Yeah I attempted the question last night and I did get an answer that I think is reasonable - not sure if it's correct though.
So, assuming the 37.238g of amino acid dissolved evenly throughout the 1L solution of 0.1M HCl, a 50mL sample of this solution would contain 1.862g of the amino acid:
37.238g / 1L = x / 0.050L, x = (37.238g x 0.050L) / 1L = 1.862g in the 50 mL sample
NaOH required to reach the 1st equivalence point: nNaOH = 8812M x 0.0227L = 0.02mol
NaOH required to reach the 2nd equivalence point: nNaOH = 0.8812M x 0.03686L = 0.0325mol
0.0325 – 0.02 = 0.0125mol of NaOH is required to deprotonate 1 equivalent of the amino acid, meaning that there are 0.0125mol of the amino acid in the solution.
MWAmino Acid = (1.862g) / (0.0125mol) = 148.96g/mol ≈ 149g/mol
Therefore, the molecular weight of the unknown amino acid is 149g/mol.