Question
An aqueous solution that is 30.0 percent sulfuric acid (H2SO4) by mass has a density of 1.105 g/mL. Determine the molarity of the solution.
Answers
water is 1 g/mL
so we have 0.105 g of H2SO4 per mL
How many mols is that
H2 = 2 g/mol
S = 32 g/mol
4 O = 64 g/mol
total = 98 g/mol
.105/98 = .0011 mol /mL which is
1.1 mol/Liter
so we have 0.105 g of H2SO4 per mL
How many mols is that
H2 = 2 g/mol
S = 32 g/mol
4 O = 64 g/mol
total = 98 g/mol
.105/98 = .0011 mol /mL which is
1.1 mol/Liter
I disagree with Damon's answer.
1.105 g/mL x 1000 mL x 0.30 x (1 mol/98g) = 3.4 M
1.105 g/mL x 1000 mL x 0.30 x (1 mol/98g) = 3.4 M
I might make that 3.38 M to 3 s.f.
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