Two equal-volume boxes illustrate aqueous Na₂SO₄ solutions
CHEMISTRY · Single correct · AP · Chemistry · Properties of Substances and Mixtures · Properties of Substances and Mixtures
Problem
Two equal-volume boxes illustrate aqueous Na₂SO₄ solutions.
• Box X (labeled “dilute”) contains 8 Na⁺ ions, 4 SO₄²⁻ ions, and 80 H₂O molecules.
• Box Y (labeled “concentrated”) contains 4 Na⁺ ions, 2 SO₄²⁻ ions, and 60 H₂O molecules.
Assuming the diagrams are otherwise correct, which adjustment would correctly represent the relative concentrations of the two solutions?
(a) Keep the current labels, because Box Y has fewer water molecules than Box X and therefore represents the higher concentration.
(b) Add four Na+ ions and two SO42− ions to Box Y so that its solute-to-solvent ratio exceeds that of Box X, justifying the original labels.
(c) Relabel Box X as “concentrated’’ and Box Y as “dilute’’ because Box X has the greater ratio of solute ions to water molecules.
(d) Relabel both boxes as having the same concentration, because each shows the proper 2:1 ratio of Na+ to SO42− ions.