A sample of an alkali XOH of mass 2.8 g was dissolved in water.

This solution was neutralised by 12.5 cm3 of sulphuric acid of concentration 2.0 mol dm-3
.
2 XOH (aq) + H2SO4 (aq) → X2SO4 (aq) + 2 H2O (l)
What is the relative atomic mass of X?
(Ar : H = 1, O = 16, S = 32)

Answer:

26) C

Volume of sulphuric acid = 12.5 = 0.0125 dm³
Concentration of sulphuric acid = 2 mol/dm³
Moles of sulphuric acid = concentration x volume = 0.0125 x 2 = 0.025 mol

Molar ratio (XOH: H₂SO₄) = 2:1
Moles of XOH = 0.025 x 2 = 0.05 mol
Mass of XOH = 2.8g ≈ 3

Mr of XOH = mass ÷ moles = 3 ÷ 0.05 = 60
Ar of H = 1
Ar of O = 16
Ar of X = 60 – 1 – 16 = 43

This is closest to 39. So C is the answer.