Q52 / 180 · AIIMS 2019 (25 May) · Chemistry
An element germanium crystallises in $b c c$ type crystal structure with edge of unit cell equal to $288 \mathrm{pm}$ and the density of the element equal to $7.2 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of atoms present in $52 \mathrm{~g}$ of the crystalline element and the atomic mass of the element are respectively
- A$2.4053 \times 10^{27}, 51.8$
- B$3.023 \times 10^{23}, 48.5$
- Correct answerC$6.046 \times 10^{23}, 51.8$
- D$6.023 \times 10^{23}, 60.0$
Solution
$d=7.2 \mathrm{~g} \mathrm{~cm}^{-3}, a=288 \mathrm{pm}$ and mass of element $=52 \mathrm{~g}$
Hence, $d=\frac{\text { Mass }}{\text { Volume }} \Rightarrow 7.2 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{52 \mathrm{~g}}{V}$
$V=\frac{52 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}=7.222 \mathrm{~cm}^3$
Edge of unit cell, $a=288 \mathrm{pm}=288 \times 10^{-12} \mathrm{~m}$ $=288 \times 10^{-10} \mathrm{~cm}=2.88 \times 10^{-8} \mathrm{~cm}$
Volume of unit cell, $a^3=\left(2.88 \times 10^{-8} \mathrm{~cm}\right)^3$ $=23.888 \times 10^{-24} \mathrm{~cm}^3$
The number of unit cells in $52 \mathrm{~g}$ of metal
$\begin{aligned} & =\frac{\text { Total volume }}{\text { Volume of one unit cell }} \\ & =\frac{7.222 \mathrm{~cm}^3}{23.888 \times 10^{-24} \mathrm{~cm}^3}=0.3023 \times 10^{24}\end{aligned}$
Germanium has body centred cubic structure. Hence, its unit cell contains 2 atoms.
$\therefore$ Total atoms in $3.023 \times 10^{23}$ unit cells are $2 \times 3.023 \times 10^{23}=6.046 \times 10^{23}$ atoms
Molar mass $=\frac{52 \mathrm{~g} \times 6.022 \times 10^{23} \text { atom } \mathrm{mol}^{-1}}{6.046 \times 10^{23} \text { atom }}$
Hence, atomic mass $=51.8$
Hence, the correct answer is option: 'C'.