0.01 m aqueous formic acid solution freezes at – 0.021°C. Calculate its degree of dissociation. (Kf = 1.86 K kg mol–1)(b) Write properties of interstitial compounds.

Answer: (a) α ≈ 0.129 (12.9%)

Observed \(\Delta T_f = 0 - (-0.021) = 0.021\) K. Calculated (no dissociation) \(\Delta T_f = K_f m = 1.86 \times 0.01 = 0.0186\) K.

van’t Hoff factor \(i = \dfrac{0.021}{0.0186} = 1.129\)

HCOOH ⇌ HCOO^– + H⁺, so one molecule gives \(n = 2\) particles: \(i = 1 + \alpha(n-1) = 1 + \alpha\).

\(\alpha = i - 1 = 1.129 - 1 = 0.129\), i.e. formic acid is about 12.9% dissociated.

(b) Properties of interstitial compounds (formed when small atoms like H, B, C, N occupy the voids in the lattice of transition metals, e.g. TiC, Mn₄N, Fe₃H):

• They have higher melting points than the pure metals.
• They are very hard; some borides are nearly as hard as diamond.
• They retain metallic conductivity and lustre.
• They are chemically inert and usually non-stoichiometric (neither typically ionic nor covalent).