Chemistry problem help please?
The question is:
What is the mole fraction of sulfuric acid in a solution made by adding 3.4 grams of sulfuric acid to 3,500 mL of water?
I need a step-by-step explanation and answer please!! Thank you so much for any help!! (:
- Anonymous7 years agoFavorite Answer
Firstly the mole fraction is just a way of defining the composition of a mixture so how much of each thing is in the mixture in (mols/mol)
First you need to find how many moles of sulphuric acid you have:
moles = mass/Mr (Mr of H2SO4 = 98); moles = 3.4/98 = 0.0347 moles of sulphuric acid
3500 mL is equal to 3500 g of water, because the 1 mL of water = 1 g of water - they have the same density.
Now we can find the moles of water:
moles = mass / Mr (Mr H2O = 18) moles = 3500/18 = 194.4 moles
now we can add the number of moles together = 0.0347 +194.4 = 194.4347
moles fraction = 0.0347/ 194.4347 = 1.80 x 10^-4
1/ 1.80 x 10^-4 = 5603.3 mols/mol
- 7 years ago
mole fraction is mol of A/( total moles)
1) find moles of sulfuric acid
4.5 g x (1 mole H2SO4/98.0g/mol) = 0.0459 mol H2SO4
2) find moles of water
3500mLx ( 10^-3/1L) x (1 L / 1.0 g) x (1 mole H2O /18g H2O) =0.194 mol H2O
3) find the mole fraction of sulfuric acid
mole fraction of H2SO4 =( 0.0459) /( 0.0459 +0.194) = 0.191Source(s): chemistry student