Q21 / 160 · TS EAMCET 2024 (10 May Shift 1) · Chemistry

An element with molar mass $2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}$ forms a cubic unit cell with edge length of 405 pm . If its density is $2.7 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, the number of atoms present in one unit cell of it is (Given : $\left.\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} \mathrm{~mol}^{-1}\right)$

  • A
    2
  • Correct answer
    B
    4
  • C
    6
  • D
    12

Solution

The number of atoms present in one unit of element can be calculated by $\begin{aligned} & n=\frac{N_a \times \rho}{M \times V} \\ & n=\text { Number of atom per unit } \\ & \rho=\text { density of element } \\ & N_2=6.023 \times 10^{23} \\ & M=\text { Molar mass of element }=2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1} \\ & V=\text { Volume of unit cell } \\ & V=a^3=\left(405 \times 10^{-12}\right)^3=6.685 \times 10^{-29} \\ & n=\frac{6.023 \times 10^{23} \times 2.7 \times 10^3}{2.7 \times 10^{-2} \times 6.685 \times 10^{-29}} \\ & n=4 \end{aligned}$
Number of atom per unit cell of the element is approximately 4.

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

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