[moni]

1.   A 1.80g sample of acid H2X, required 14.00 ml of KOH solution for neutralization of all the hydrogen ions. Exactly 14.2 ml of this same KOH solution was found to neutralize 10.0 ml of 0.750 M H2SO4 .Calculate the molecular mass of H2X?

2.  What mass of sulphuric acid solution (95% by mass) is needed to prepare 200.0g of a 20.0% solution of the acid by mass?

THANX FOR HELP

[Donaldson Tan]

here's a hint: write the material (mass) balance for question (1).

btw please show that you had attempted the problem.

i will lock this topic in 3days time if no attempt is shown by the threadstarter.

[moni]

 yes i try ,the equation for this reaction 2KOH + H2SO4-->K2SO4 +2H2O   ; AND FOR THE OTHER 2KOH +H2X---> K2X+2H2O

FROM M1V1=M2V2 = 0.528M KOH
 
AND THAN I DONT KNOW

Mol KOH/L of soln * L of Koh requied= mol Koh consumed   ?

mole Koh consumed*1mol H2X/ 2mol KOH= mol H2X present

thank you for help

[Albert]

FROM M1V1=M2V2 = 0.528M KOH

Not at all. The chemical equations you wrote are both correct. You should see which is the ratio H2SO4/KOH.

[Borek]

AND THAN I DONT KNOW

Fix the CapsLock.

[Albert]

Mol KOH/L of soln * L of Koh requied= mol Koh consumed   ?

mole Koh consumed*1mol H2X/ 2mol KOH= mol H2X present

That's true: you are on the right track. What you need now is to determine KOH molarity and, after the calculations you described, use the definition of molecular weight in order to solve your problem.