2.9 · SolutionsPractice · 溶液

Hexane is a nonpolar solvent. “Like dissolves like”: nonpolar solutes dissolve in nonpolar solvents, while ionic and polar solutes dissolve in polar solvents such as water. Sodium chloride is ionic; sucrose is polar; naphthalene is nonpolar; iodine is nonpolar. So naphthalene and iodine dissolve well in hexane, while sodium chloride and sucrose dissolve in water but hardly at all in hexane.

Dissolving means pulling the ions out of the NaCl crystal. Water is polar: the partially negative O end of each molecule faces $\ce{Na+}$ and the partially positive H end faces $\ce{Cl-}$, surrounding each ion (hydration). That attraction makes up for the energy needed to break up the crystal. Gasoline is made of nonpolar molecules with almost no attraction to ions. No new substance forms when NaCl dissolves, NaCl is ionic, and density has nothing to do with solubility.

n = M × V = 0.150 mol/L × 0.250 L = 0.0375 mol (convert mL to L). Mass = 0.0375 mol × 105.99 g/mol = 3.97 g. Dissolve it and dilute to the mark in a 250-mL volumetric flask, rather than adding 250 mL of water.

An ion’s concentration = the solute’s concentration × the number of those ions per formula unit; the volume doesn’t matter. In $\ce{AlCl3}$, [Cl⁻] = 0.30 × 3 = 0.90 M. The choices: $\ce{NaCl}$: 0.30 × 1 = 0.30 M; $\ce{CaCl2}$: 0.45 × 2 = 0.90 M; $\ce{MgCl2}$: 0.30 × 2 = 0.60 M; $\ce{KClO3}$: it gives $\ce{ClO3-}$, not $\ce{Cl-}$.

Dilution doesn’t change the moles of solute: M₁V₁ = M₂V₂. V₁ = 0.600 × 250 ÷ 12.0 = 12.5 mL. Add it to some water, then dilute to the mark in a 250-mL volumetric flask.

Find the moles of Cl⁻ from each, then divide by the total volume. $\ce{NaCl}$: 0.200 × 1 × 0.0500 = 0.0100 mol; $\ce{CaCl2}$: 0.100 × 2 × 0.0500 = 0.0100 mol. That’s 0.0200 mol in 0.1000 L, so [Cl⁻] = 0.200 M. Remember that each $\ce{CaCl2}$ gives 2 $\ce{Cl-}$.

n(MgCl₂) = 0.150 M × 0.0200 L = 0.00300 mol. NaOH is in excess, so MgCl₂ limits: n(Mg(OH)₂) = 0.00300 mol, and m = 0.00300 × 58.33 = 0.175 g.

n(NaOH) = 1.00 ÷ 40.00 = 0.02500 mol. HCl + NaOH → NaCl + H₂O reacts 1 : 1, so n(HCl) = 0.02500 mol. V = n ÷ M = 0.02500 ÷ 0.200 = 0.1250 L = 125 mL.

Think in terms of M = n ÷ V. Dissolving NaOH releases heat (about 44.5 kJ per mole). If the warm solution is filled to the mark, it shrinks as it cools and ends up below the mark: V is too small and the concentration too high. Not rinsing leaves solute behind (n too small, concentration too low). Looking up at the mark from below means filling past it (V too large, concentration too low). Water already in the flask doesn’t change the final volume, so it has no effect.

Zinc is in excess, so the acid limits: n(HCl) = 2.00 × 0.0400 = 0.0800 mol. The equation gives 1 mol H₂ per 2 mol HCl, so n(H₂) = 0.0400 mol and V = 0.0400 × 22.4 = 0.896 L.