Problem: Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H2SO4, for which the density is 1.3057 g/mL.

$molarity (M) = \frac{moles solute}{L solution}$

mass percent of H₂SO₄ = 40.0%

$% Mass = \frac{mass of {CH}_{3} COOH}{mass of solution}$

This means in 100 g of solution, we have 40.0 g H₂SO₄

$40.0 % = \frac{40.0 g H_{2} {SO}_{4}}{100 g solution}$

molar mass H₂SO₄ = 98.079 g/mol

The moles of H₂SO₄ is:

$moles H_{2} {SO}_{4} = 40.0 g H_{2} {SO}_{4} \times \frac{1 mol H_{2} {SO}_{4}}{98.079 g H_{2} {SO}_{4}}$

moles H₂SO₄ = 0.4078 mol

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Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H₂SO₄, for which the density is 1.3057 g/mL.

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Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.