Question

An analytical chemist weighs out 0.274 g of an unknown diprotic acid into a 250 ml volumetric flask and dilutes to the mark with distilled water. She then titrates this solution with 0.0600 M NaOH solution. When the titration reaches the equivalence point, the chemist finds she has added 68.1 ml of NaOH solution. Calculate the molar mass of the unknown acid. Round your answer to 3 significant digits. mol X2 ?

Answer 1

Answer:

Given that maintain dissolved into 250 m2 distived water by we have to calculate the molar mass of 0.274 g of unknown diprofic acid volume ining To titrate it, 68.1 mhe of 0.0600 M of Nash the unknown compound. xet, the molar mass is Mol. solution required. M Then, glu 0.274 g M g. yani 9749 = 0.274 M wol So, 250 ML contains 0.274 wol M Then 100 mL or 14 -0.274 х 1000 mother 250 amol 1.096 mol lie the molarity 1.096 molicit 1096 M. (m)

we know that, [visi = V2 S2 250 ma 1.096 X (ar) = 68.1 m2 x 0.060044 = M 250X 1.096 68.14 0.0600 > M = 67.06
As it is a diprotic acid, so the required molar mass is (2×67.06) g/mol =134.12 g/mol.

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