Question

A solution of dibasic acid H2X contains 12.60gr of acid per litre, 10cm^3 of this acid solution required 20cm^3 of 0.1M potassium hydroxide solution for complete neutralization using phenolphthalein
Find relative molecular mass of of acid and deduce the atomic mass of X in formula H2X

Answer 1

H2X + 2KOH ==> K2X + 2H2O
mols = M x L = 0.1 M x 0.020 L = 0.002 mols KOH
Looking at the equation you know 1 mol H2X = 2 mols kOH; therefore, 2 moles KOH = 1 mol H2X = 0.001 mol H2X in 10 cc of the solution or 0.001 x 1000/10 = 0.1 mol H2X/L and that = 12.6 g H2X so 12.6 g H2X = 0.1 mol.
moles = grams/molar mass or
molar mass = grams/mols = 12.6/0.1 = 126 = molar mass H2X.
From that we know H is 2 for the two so X must be 124. I don't know of any element with atomic mass 124. You might hazard a guess at Te with atomic mass 127.6 which would make H2X = about 130.The only problem I have with that is that the accuracy of a titration like that is FAR BETTER than that. So I think there must be something wrong with the problem or the post has a typo. I'm a shaggy analytical chemist if you give me a sample of H2X with a molar mass of 130 and I can't do better than 126 when I titrate it.

Answer 2

so can you know that hiden element in that dibasic acid