Asked by tyler
An aqueous solution that is 10.0 percent sulfuric acid (H2SO4) by mass has a density of 1.143 g/mL. Determine the molality of the solution. ( 3 significant figures)
Answers
Answered by
DrBob222
Please don't change screen names.
m = mols/kg solvent
mols = grams/molar mass.
1.143 g/mL x 1000 mL = 1143 g = mass solution
10.0% of that is H2SO4 = 1143 x 0.100 = 114.3 g is H2SO4.
mols H2SO4 = grams/molar mass = ?
mass H2O = 1143-114.3 = 1028.7
Then m = mols/kg = ?
You take care of the number of significant figures.
m = mols/kg solvent
mols = grams/molar mass.
1.143 g/mL x 1000 mL = 1143 g = mass solution
10.0% of that is H2SO4 = 1143 x 0.100 = 114.3 g is H2SO4.
mols H2SO4 = grams/molar mass = ?
mass H2O = 1143-114.3 = 1028.7
Then m = mols/kg = ?
You take care of the number of significant figures.
Answered by
Purplehyacinth01
Arigato
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