Chemistry - Concentration Terms Question with Solution | TestHub

The 109% label means 100 g oleum + 9 g water $\to$ 109 g H₂SO₄. The 9 g (0.5 mol) of water converts free SO₃ to H₂SO₄, so free SO₃ = 0.5 mol = 40 g per 100 g sample: (a) is true. Total acidity = 109/98 = 1.1122 mol H₂SO₄ per 100 g = 2.2245 g-eq per 100 g. (b) 1 g needs 0.0222 g-eq of NaOH $\to$ 0.0222/0.5 = 44.5 ml of 0.5 M NaOH, not 22.25 ml - false. (c) 5 g needs 0.1112 g-eq $\to$ 0.1112/0.1 = 1112 ml of 0.1 N Ba(OH)₂, not 11.12 ml - false. (d) 100 g sample gives 1.1122 mol H₂SO₄ in 500 g (0.5 kg) of water $\to$ 2.224 m - true on the usual convention that the added water is the solvent.