Chemistry

Chemistry

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+\mathrm{Na^+} ions and two SO42−\mathrm{SO_4^{2-}} 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 ⁣: ⁣12\!:\!1 ratio of Na+\mathrm{Na^+} to SO42−\mathrm{SO_4^{2-}} ions.

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