Answer to Question #320209 in General Chemistry for Aljonny

Question

A 2 L sulfuric acid solution containing 320 g of H2SO4 per liter of solution has a density of 1.329 g/ mL. What is the mole fraction of the solvent in the solution? Assume that the solvent is water. 

Answer 1

V(solution) = 2 L = 2000 mL;

m(H₂SO₄) = 320 g;

d(solution) = 1.329 g/mL;

M(H₂SO₄) = 98 g/mol;

M(H₂O) = 18 g/mol;

m(solution) = V(solution, mL) * d(solution) = 2000 * 1.329 = 2658 g;

m(solvent, H₂O) = m(solution) - m(H₂SO₄) = 2658 - 320 = 2338 g;

n(H₂SO₄) = m(H₂SO₄)/M(H₂SO₄) = 320/98 = 3.27 mol;

n(solvent, H₂O) = m(solvent, H₂O)/M(solvent, H₂O) = 2338/18 = 129.89 mol;

X(solvent, H₂O) = n(solvent, H₂O)/(n(solvent, H₂O) + n(H₂SO₄)) = 129.89/(129.89+3.27) = 129.89/133.16 = 0.975 = 97.5 %.

Answer: 97.5 %.