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Stoichiometry and solutionsNational 2025

Which procedures will result in a 1.000 m solution of sodium acetate?

I. Adding 1.000 mol sodium acetate trihydrate to a 1.000-liter volumetric flask, adding deionized water to dissolve, and then filling to the mark

II. Mixing 1.000 mol sodium hydroxide with 1.000 mol acetic acid and 1.000 kg deionized water

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Explanation

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Molality is defined as $\text{mol solute} / \text{kg solvent}$ — it requires knowing the mass of solvent (water), not the volume of solution.

Procedure I: Dissolving 1.000 mol sodium acetate trihydrate, $\ce{NaC2H3O2 \cdot 3H2O}$ , in a 1.000 L volumetric flask and diluting to the mark gives a solution whose total volume is 1.000 L. This fixes the molarity at 1.000 M, not the molality. The mass of water actually present is unknown (it's whatever amount of water was needed to reach 1.000 L, minus the water of hydration already contributed by the trihydrate), so the molality is not 1.000 m.

Procedure II: Mixing 1.000 mol $\ce{NaOH}$ with 1.000 mol $\ce{CH3COOH}$ in 1.000 kg of water triggers the neutralization
$\ce{NaOH + CH3COOH -> NaC2H3O2 + H2O}$
This reaction produces additional water (about 18 g, from 1.000 mol). The solvent at the end is therefore 1.000 kg (added) + water generated by the reaction, which is more than 1.000 kg. Since molality uses the mass of solvent present, the final water mass exceeds 1.000 kg, so the molality is slightly less than 1.000 m, not exactly 1.000 m.

Since neither procedure yields exactly 1.000 mol solute per 1.000 kg solvent, the correct choice is (D) Neither I nor II.

Incorrectly assumes that diluting a known number of moles to a fixed volume (1.000 L) is equivalent to controlling molality; this actually fixes the molarity, not the molality, and ignores the unknown water mass.

Assumes that mixing 1.000 mol acid with 1.000 mol base in 1.000 kg water gives exactly 1.000 m sodium acetate, overlooking that the neutralization reaction itself generates extra water as a product, increasing the true solvent mass beyond 1.000 kg.

Combines both errors above, mistakenly believing that fixing solution volume (I) and fixing only the added water mass while ignoring water produced by reaction (II) both correctly control molality.

Correct: molality depends on the mass of solvent present. In (I) the solvent mass is unknown (only total volume is fixed), and in (II) the reaction produces additional water, so in neither case is the solvent mass exactly 1.000 kg per 1.000 mol solute.

Concept · Molality vs. molarity; water produced in neutralization reactions