MHTCET 15 Apr 2026 Shift 1 Paper

Concept:For an fcc unit cell, the number of atoms per unit cell is $Z = 4$.The relation between density, atomic mass, and cell edge length is used to find the molar mass.Explanation:The density formula for a unit cell is$$\rho = \frac{Z \times M}{a^{3} \times N_A}$$Rearranging to solve for atomic mass $M$,$$M = \frac{\rho \times a^{3} \times N_A}{Z}$$Given values are$\rho = 8.92\ \mathrm{g\ cm}^{-3}$, $a = 3.608 \times 10^{-8}\ \mathrm{cm}$, and $N_A = 6.022 \times 10^{23}\ \mathrm{mol}^{-1}$.First calculate $a^{3}$:$$a^{3} = \left(3.608 \times 10^{-8}\right)^{3} = 4.697 \times 10^{-23}\ \mathrm{cm}^{3}$$Substitute the values into the formula:$$M = \frac{8.92 \times 4.697 \times 10^{-23} \times 6.022 \times 10^{23}}{4}$$Simplify the powers of ten and calculate:$$M = \frac{8.92 \times 4.697 \times 6.022}{4} \approx \frac{252.3}{4} \approx 63.1\ \mathrm{g\ mol}^{-1}$$Thus, the atomic mass of the element is approximately $63\ \mathrm{g\ mol}^{-1}$.Answer:The atomic mass of the element is $\mathbf{63\ g\ mol^{-1}}$.Therefore, the correct option is Option C ($63\ \mathrm{g/mol}$).