Assam ASSEB HS Chemistry study materials
Assam ASSEB HS Chemistry study resources

Assam ASSEB HS Chemistry: Solid State and Solutions

Assam ASSEB HS Chemistry · Work through your exam, one topic at a time.

All guide sections

Solid State and Solutions

What this domain tests

Solid State examines unit-cell geometry (Z, coordination number, packing efficiency, radius–edge relations), density calculation and point defects. Solutions examines concentration terms, Raoult's law, ideal and non-ideal behaviour, Henry's law and the four colligative properties with the van't Hoff factor.

Core rules

  • Packing efficiency: simple cubic 52.4% (CN 6, Z = 1), bcc 68% (CN 8, Z = 2), fcc/hcp 74% (CN 12, Z = 4).
  • Radius relations: fcc a = 2√2 r; bcc a = 4r/√3.
  • Density d = ZM/(a³N_A); convert pm to cm by multiplying by 10⁻¹⁰ before cubing.
  • Schottky defect removes equal numbers of cations and anions and lowers density; Frenkel defect moves a cation to an interstitial site and leaves density unchanged.
  • ΔTb = iKb·m, ΔTf = iKf·m, π = iCRT (C in mol L⁻¹, R = 0.0821 L atm K⁻¹ mol⁻¹).
  • Relative lowering of vapour pressure equals the mole fraction of solute for a non-volatile solute.

Worked example

Copper crystallises in fcc with a = 361 pm and M = 63.5 g mol⁻¹. Here Z = 4, so a³ = (3.61 × 10⁻⁸ cm)³ = 4.71 × 10⁻²³ cm³. Then d = (4 × 63.5)/(4.71 × 10⁻²³ × 6.022 × 10²³) = 254/28.4 = 8.9 g cm⁻³. The value matches the measured density, confirming the fcc assignment.

Traps that cost marks

Colligative formulae need molality, not molarity, for ΔTf and ΔTb. i > 1 for dissociation and i < 1 for association; benzoic acid in benzene gives i close to 0.5. For a mixture of two volatile liquids, apply Raoult's law to both components. Do not interchange Schottky and Frenkel descriptions.