[UG]

Hey, for this problem: How many liters of water are needed to prepare a 45% (by mass) solution of sulfuric acid from 50 kg of 35% (by mass) oleum.
What I did was, calculated the m(H₂SO₄) = 32500 g and m(SO₃) = 17500 g
So 32500 g now has to be 45% of the mass, so I divided 32500 / 0.45 = 72222 g, so the mass of water = 72222-50000 = 22222 g which is ~ 22.2 L. But this isn' the correct answer  Where did I go wrong please someone?

[Borek]

What is oleum?

[UG]

Oleum is sulfur trioxide in sulfuric acid, kind of like H₂SO₄.xSO₃

[Borek]

So, what happens when you add water?

[UG]

How stupid I am. Thanks.

[UG]

I've got one more query, when they say 35% oleum by mass, do they mean for example, in 100 gram, there is 35 g of SO₃ and 65 grams of H₂SO₄ OR do they mean that there is 65 grams of H₂SO₄ and 35 grams of H₂SO₄.SO₃? 

[Borek]

From what I remember 35% SO₃.