Two Unlabelled White Solids and a Very Hard Third
Two unlabelled white solids are found in a school store room. Solid A melts at about 801 °C; it does not conduct electricity as a solid but conducts well when molten. It dissolves readily in water, the resulting solution conducts, and a flame test on A gives an intense yellow-orange colour. Solid B melts at 186 °C; it does not conduct as a solid or when molten, and although it dissolves in water the solution does not conduct. Adding a few drops of concentrated sulfuric acid to B chars it to a black mass. A third sample, C, is a hard, brittle, colourless solid that melts above 1,600 °C and does not conduct even when molten.
The teacher asks the class to explain the behaviour of A, B and C in terms of structure and bonding, and then to predict what happens when 5.85 g of A and 34.2 g of B are each dissolved in separate 500 g samples of water.
1. Classify each solid by bonding type and justify the classification from every observation given.
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Solid A - ionic lattice. The high melting point (801 °C) reflects strong electrostatic attraction in the lattice, not discrete molecules. The conduction pattern is diagnostic: mobile ions are needed to carry current, and in the solid the ions are locked in place, so it insulates; melting frees them, so the melt conducts. Aqueous conduction shows the same ions in solution. The intense yellow-orange flame is the emission from excited sodium atoms, so A is almost certainly sodium chloride.
Solid B - molecular covalent. A low melting point (186 °C) means weak intermolecular forces are being overcome, not covalent bonds. No conduction in either state is expected because there are no free ions or delocalised electrons. The aqueous solution is non-conducting, which rules out an ionic solid outright - the molecules disperse but do not dissociate into ions. Charing with concentrated sulfuric acid indicates a carbohydrate/organic molecule (dehydration to carbon); sucrose fits all the data.
Solid C - network covalent (giant covalent lattice). Extreme melting point, hardness and brittleness are the signature of a continuous three-dimensional covalent network, and the absence of conduction even when molten rules out a metal or an ionic compound. Fused silica or a similar network solid fits.
Key discipline: no single observation settles the assignment; it is the combination of melting point, conduction in three states, and the solution behaviour that separates the three categories.
2. Predict and calculate the boiling-point elevation for each solution (Kb for water = 0.512 °C kg mol⁻¹). Explain the van't Hoff factor and comment on whether the ideal value is realistic.
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Amounts: n(A) = 5.85 g ÷ 58.5 g mol⁻¹ = 0.100 mol; n(B) = 34.2 g ÷ 342 g mol⁻¹ = 0.100 mol. Each in 0.500 kg water gives molality = 0.100/0.500 = 0.200 mol kg⁻¹. The molalities are deliberately identical, so any difference in effect is purely about particle number.
Calculations:
- Solution B (sucrose, i ≈ 1): ΔTb = i·Kb·m = 1 × 0.512 × 0.200 = 0.102 °C, boiling near 100.10 °C.
- Solution A (NaCl, i = 2 ideal): ΔTb = 2 × 0.512 × 0.200 = 0.205 °C, boiling near 100.21 °C.
Why i = 2: colligative properties count dissolved particles, not formula units. Each NaCl unit dissociates into Na⁺ and Cl⁻, so 0.200 mol kg⁻¹ of salt supplies 0.400 mol kg⁻¹ of ions.
Is i = 2 realistic? No - it is the upper limit. In real solution, ion pairing (Na⁺ and Cl⁻ transiently associated, solvent-shared) means the effective i for 0.2 m NaCl is closer to 1.85–1.90, giving ΔTb ≈ 0.19 °C. The ideal value overestimates the effect, and the discrepancy grows with concentration. Sucrose, being a non-electrolyte, has no such correction.
3. Predict which of the two solids would dissolve in hexane, and explain the reasoning. Then explain why the freezing-point effect is roughly twice as large for A as for B.
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Hexane test: hexane is non-polar, with only dispersion forces available. Sucrose (B) presents numerous O–H groups and is strongly hydrogen-bonding and highly polar; the energetic cost of breaking sucrose–sucrose hydrogen bonds is not repaid by any favourable interaction with hexane, so it is essentially insoluble. Sodium chloride (A) is worse still: dissolving it requires separating Na⁺ and Cl⁻ from the lattice, and a low-permittivity solvent like hexane cannot solvate the ions or screen their charges. Ionic solids do not dissolve in non-polar solvents. Neither dissolves appreciably in hexane - the honest answer - though the reason differs: lattice energy versus hydrogen bonding.
Freezing point: ΔTf = i·Kf·m, with Kf(water) = 1.86 °C kg mol⁻¹. At 0.200 m: sucrose gives 1 × 1.86 × 0.200 = 0.372 °C depression; NaCl gives 2 × 1.86 × 0.200 = 0.744 °C ideally. The factor of two comes entirely from dissociation - the same number of formula units delivers twice the number of solute particles. Measured values are slightly smaller for the salt (~0.70 °C) because of ion pairing, so the ideal ratio is approached but not reached. Both are small compared with the 1.86 °C per molal unit, which is why the effect is usually measured with a precision thermometer rather than by feel.