Answer: (a) n = 4, face-centred cubic (fcc) unit cell; (b) derivation below.
(a) \(\rho = \dfrac{nM}{a^3 N_A} \Rightarrow n = \dfrac{\rho a^3 N_A}{M}\)
\(a = 405\) pm \(= 4.05\times10^{-8}\) cm; \(a^3 = (4.05)^3 \times 10^{-24} = 66.43\times10^{-24}\) cm\(^3\)
\(n = \dfrac{2.7 \times 66.43\times10^{-24} \times 6.022\times10^{23}}{27} = 0.1 \times 40.0 = 4\)
Four atoms per unit cell, so the unit cell is face-centred cubic.
(b) Raoult's law: the vapour pressure of the solvent over a solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction: \(P_1 = P_1^0 x_1\).
For a binary solution of a non-volatile solute, only the solvent contributes to the vapour pressure, so \(P = P_1 = P_1^0 x_1\).
Since \(x_1 + x_2 = 1\), \(x_1 = 1 - x_2\):
\(P_1 = P_1^0(1 - x_2) = P_1^0 - P_1^0 x_2\)
\(\therefore P_1^0 - P_1 = \Delta P = P_1^0 x_2\) and \(\dfrac{P_1^0 - P_1}{P_1^0} = x_2\)
The relative lowering of vapour pressure equals the mole fraction of the solute. For a dilute solution \(x_2 \approx \dfrac{n_2}{n_1} = \dfrac{W_2 M_1}{M_2 W_1}\).
