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!! (:

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• Anonymous
7 years ago

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

• 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) = 0.191

Source(s): chemistry student