Q35 / 150 · MHT CET 2020 (19 Oct Shift 2) · Chemistry

An element crystallises in fcc type of unit cell. The volume of one unit cell is $24.99 \times 10^{-24} \mathrm{~cm}^{3}$ and density of the element $7 \cdot 2 \mathrm{~g} \mathrm{~cm}^{-3} .$ Calculate the number of unit cells in $36 \mathrm{~g}$ of pure sample of element?

  • Correct answer
    A
    $2 \cdot 0 \times 10^{23}$
  • B
    $2 \cdot 0 \times 10^{21}$
  • C
    $2 \cdot 0 \times 10^{24}$
  • D
    $1 \cdot 25 \times 10^{21}$

Solution

(C) Vol. of element $=\frac{\text { Mass }}{\text { Density }}=\frac{36 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}=5 \mathrm{~cm}^{3}$ No. of unit cells in $36 \mathrm{~g}$ of pure sample of element $=\frac{\text { Total Vol. of element }}{\text { Vol. of one unit cell }}$ $=\frac{5 \mathrm{~cm}^{3}}{24.99 \times 10^{-24} \mathrm{~cm}^{3}}=2.0 \times 10^{23}$

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

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