A solid 'AB' crystallizes as CCP for 'A' atoms and 'B' atoms occupy half of the tetrahedral voids. O — Solid State Chemistry Question
Question
A solid 'AB' crystallizes as CCP for 'A' atoms and 'B' atoms occupy half of the tetrahedral voids. One litre of the crystal is doped with 1 mole atoms of 'C', some of which replace the 'B' atoms and remaining occupies new interstitial voids without affecting the dimensions of the crystal. If the density of the crystal before doping is $4.8 \text{ g/cm}^3$ and the density after doping is $4.795 \text{ g/cm}^3$, the percentage of 'C' atoms which replace 'B' atoms is (Atomic masses of A, B and C are 40, 30 and 15, respectively)
💡 Solution & Explanation
Initial density = $4800 \text{ g/L}$. Molar mass of AB = $70 \text{ g/mol}$. Initially, there are $4800/70 = 68.57$ moles of AB per litre. Adding 1 mole of C (15 g). Let $x$ moles of B (30 g/mol) be replaced. Change in mass = $15 - 30x$. New density = $4795 \text{ g/L}$, meaning mass dropped by 5 g. So $15 - 30x = -5 \implies 30x = 20 \implies x = 2/3$. The percentage of C replacing B is $2/3 \times 100 \approx 66.67\% \approx 67\%$. Therefore, correct answer is 0067.