Answer to Question #344375 in Inorganic Chemistry for Cess

Question

A 0.00453 L of sodium chloride solution was needed in a 560 mg sample that assayed

87.1% (w/w) Na2SO4. Calculate the molar concentration of sodium chloride in the

solution.

Answer 1

C = n(solute) / V(solution)

V(Na₂SO₄) = m(Na₂SO₄) / ρ(Na₂SO₄) = (0.56 g * 87.1 %) / 2.66 g/cm³ = 18.34 cm³ = 0.01834 L

n(NaCl) = m(NaCl) / Mr(NaCl) = (ρ(NaCl)*V(NaCl)) / Mr(NaCl) = (2.16 g/cm³ * 4.53 cm³) / 58,44 g/mol = 0.17 moles

C = 0.17 moles / 0.01834 L = 9.27 mol/L