MHTCET 22 Apr 2025 Shift 2 Paper

Concept:In a face-centered cubic (fcc) unit cell, the total mass of the unit cell equals density times volume, and it contains 4 atoms per unit cell.Explanation:Step 1: The mass of one unit cell is given by the product of density and unit cell volume.$$M_{\text{cell}} = \rho \times a^{3} = 6.8 \times 10^{-22} \, \text{g}$$Step 2: For an fcc lattice, the number of atoms per unit cell is $Z = 4$.Step 3: Find the mass of a single atom by dividing the unit cell mass by the number of atoms in it.$$m_{\text{atom}} = \frac{M_{\text{cell}}}{Z} = \frac{6.8 \times 10^{-22}}{4} = 1.7 \times 10^{-22} \, \text{g}$$Step 4: The number of atoms in 1 g of the element is obtained by dividing 1 g by the mass of one atom.$$N = \frac{1}{1.7 \times 10^{-22}}$$$$N = 5.882 \times 10^{21}$$Thus, 1 g of the element contains $5.882 \times 10^{21}$ atoms.Answer:$$5.882 \times 10^{21}$$ atoms, which matches Option D.