Q26 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Chemistry

If each edge of a cubic unit cell of an element having atomic mass 120 and density $6.25 \mathrm{~g} \mathrm{cc}^{-1}$ measures $400 \mathrm{pm}$, then the crystal lattice is

  • A
    primitive
  • Correct answer
    B
    body centered
  • C
    face centered
  • D
    end centered

Solution

Atomic weight $(m)=120$ $N_A=6.022 \times 10^{23}$ Unit cell $(a)=400 \mathrm{pm}$ $=400 \times 10^{-10} \mathrm{~m}$ As we know $d=\frac{Z M}{N_A \times a^3}$ $6.25=\frac{Z \times 120}{6.022 \times 10^{23} \times\left(400 \times 10^{-10}\right)^3}$ $Z=2$ So, the crystal lattice is body centered.

Hence, the correct answer is option: 'B'.

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