MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Silver crystallises in a face-centered cubic (FCC) lattice. For FCC, the number of atoms per unit cell is $Z = 4$.The density of the crystal is calculated using the unit cell volume and molar mass through the standard formula.Explanation:The density formula for a crystalline solid is written as:$$< br> \rho = \frac{Z \times M}{N_A \times a^3} < br>$$Here, the values are substituted as follows:$Z = 4$, $M = 108 \ \mathrm{g\,mol^{-1}}$, $N_A = 6.022 \times 10^{23}$, and $a^3 = 6.84 \times 10^{-23} \ \mathrm{cm^3}$.Plugging these into the formula:$$< br> \rho = \frac{4 \times 108}{6.022 \times 10^{23} \times 6.84 \times 10^{-23}} < br>$$The powers of ten cancel out, giving:$$< br> \rho = \frac{432}{6.022 \times 6.84} < br>$$This simplifies to:$$< br> \rho = \frac{432}{41.19} \approx 10.49 \ \mathrm{g\,cm^{-3}} < br>$$Hence, the density of silver is $10.49 \ \mathrm{g\,cm^{-3}}$.Answer:$10.49 \ \mathrm{g\,cm^{-3}}$ — Option B.