Question

A contains 2g/200cm³ of H2X solution. B contains 2.85g/250cm³ of KOH solution. If 18.70cm³ of A neutralized 25cm³ of B. Calculate the;
i) Concentration of A in mol/dm³
ii) Molar Mass of acid H2X.
iii) Percentage by mass of X in H2X.

Answer 1

NOTE: I think you meant 2 g H2X/200 cc solution for A and 2.85 g KOH/250 cc solution for B. I will assume that's what you meant.
(A) = (H2X) = 2 g H2X/200 cc x (1000 cc/200) = 10 g H2X/L of solution.
(B) = KOH = 2.85 g KOH/250 cc x (1000 cc/250) = 11.4 g KOH/L of solution.
Find molarity of KOH. mols KOH = g/molar mass = 11.4/56.1 = 0.203 mols/L = 0.203 M for KOH.
The reaction is H2X + 2KOH ==> K2X + 2H2O
cc KOH = 25.0 cc. cc H2X = 18.70 cc.
millimols KOH = mL x M = 25.0 x 0.203 = 5.08
millimols H2X = 1/2 that or 5.08/2 = 2.54 (because it's 1 mol H2X for 2 mols KOH. millimols H2X = mL x M or M = millimols/mL = 2.54/18.70 = 0.135 M for (H2X) = 0.135 mol/dm3
mols H2X = grams/molar mass or molar mass = grams/mols = 10/0.135 = 74.1
(iii) molar mass H2X = 74.1. H = 2*1 = 2 and 74.1 - 2 = 72.1 for X
%X in H2X = (72.1/74.1)*100 = ?
Please check the math thoroughly. I made a couple of math error but I think I caught and corrected all of them. For whatever it's worth I don't know of an anion that has an atomic mass of 72.1 so I assume this is a made up problem.