Solid State and Solutions
This domain is examined as two linked numerical clusters: unit-cell and defect reasoning, then concentration and colligative-property calculations.
Unit cells and packing
- fcc/ccp: Z = 4, coordination number 12, packing efficiency 74%, r = a/(2√2)
- bcc: Z = 2, CN 8, 68%, r = √3a/4
- simple cubic: Z = 1, CN 6, 52.4%
- density d = ZM/(a³N_A); convert pm to cm before substituting
Point defects
- Schottky: a cation–anion pair is missing; density falls (NaCl, KCl)
- Frenkel: a cation moves to an interstitial site; density unchanged (AgCl, ZnS)
- F-centre: an electron trapped in an anion vacancy gives colour, e.g. yellow NaCl on heating in sodium vapour
Solution rules
- Raoult's law: p_A = x_A p°_A; ideal solutions obey it over the whole range and show zero enthalpy of mixing
- azeotropes (ethanol–water 95%) cannot be separated by fractional distillation
- colligative properties: ΔTb = i·Kb·m, ΔTf = i·Kf·m, π = iCRT
- van't Hoff factor i > 1 for dissociation, i < 1 for association (benzoic acid in benzene, i ≈ 0.5)
Worked example
Copper is fcc with a = 361 pm and M = 63.5. Then d = 4 × 63.5 / [(3.61 × 10⁻⁸ cm)³ × 6.022 × 10²³]. Now a³ = 4.706 × 10⁻²³ cm³, so the denominator is 4.706 × 10⁻²³ × 6.022 × 10²³ = 28.34. Hence d = 254/28.34 = 8.96 g cm⁻³.
Traps that cost marks
- Substituting a in pm directly into the density formula.
- Taking i = 1 for a salt; KCl gives i = 2.
- Claiming Frenkel defects change density.
- Forgetting that azeotropic composition changes when external pressure changes.