U3.7 · Solutions and MixturesPractice · 溶液与混合物

The water of hydration counts in the molar mass: M = 249.69 g/mol. n = 2.50 ÷ 249.69 = 0.01001 mol, and each $\ce{CuSO4.5H2O}$ gives one $\ce{Cu^2+}$, so $[\ce{Cu^2+}]$ = 0.01001 ÷ 0.1000 = 0.100 M. Using the molar mass of $\ce{CuSO4}$ (159.61) gives 0.157 M, too high.

Dilution doesn’t change the moles of solute: M₁V₁ = M₂V₂, so M₂ = 0.480 × 25.0 ÷ 200.0 = 0.0600 M. $\ce{KNO3}$ dissociates completely, one $\ce{K+}$ per formula unit, so $[\ce{K+}]$ = 0.0600 M. Note that 200.0 mL is the final total volume, not the volume of water added.

n($\ce{Na+}$) = 0.0500 × 0.200 × 3 + 0.0250 × 0.300 = 0.0300 + 0.0075 = 0.0375 mol. In 0.0750 L, $[\ce{Na+}]$ = 0.0375 ÷ 0.0750 = 0.500 M. Each $\ce{Na3PO4}$ gives 3 $\ce{Na+}$.

A solution is a homogeneous mixture: the same composition throughout, with no visible boundaries. Air (a gaseous solution), brass (a solid solution of copper and zinc), and vinegar (acetic acid in water) are all homogeneous. Sand doesn’t dissolve and settles out, so the composition varies from place to place: a heterogeneous mixture.

Total ion concentration = solute concentration × ions per formula unit: $\ce{Na3PO4}$ 0.10 × 4 = 0.40 M, $\ce{MgCl2}$ 0.15 × 3 = 0.45 M, $\ce{KNO3}$ 0.20 × 2 = 0.40 M, $\ce{CH3OH}$ 0.30 × 0 = 0.00 M. $\ce{CH3OH}$ is a nonelectrolyte and doesn’t ionize; the most concentrated solution isn’t necessarily the one with the most ions.