Find out the % labelling of oleum Sulphate in which mole fraction of SO3 is 0.2 ?.

SOLUTION: We know mole mass fraction percentage is equal to % labelling.

9 gm water is added into Oleum sample labelled as 112% H2SO4 then the amount of free SO3 remaining in the solution is ? (STP=1atm and 273K).

SOLUTION: Initial free moles of SO₃=  =2/3 moles

Moles of water that combined with free moles of SO₃=9/18=1/2 moles

Moles of free SO₃ left 2/3-1/2=1/6 moles

Volume of free SO₃ at STP=1/6X22.4=3.73 L