Q33 / 180 · KCET 2024 · Chemistry
The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is
(Given : Density $=10 \mathrm{~g} \mathrm{~cm}^{-3}$ and
$\left.N_A=6.022 \times 10^{23}\right)$
- A$6.6 \times 10^{20}$
- B$6.6 \times 10^{23}$
- C$6.6 \times 10^{19}$
- Correct answerD$6.6 \times 10^{22}$
Solution
Given, Face centred crystal, so $Z=4$
edge length $(a)=300 \mathrm{pm}=300 \times 10^{-10} \mathrm{~cm}$
density $(d)=10 \mathrm{~g} / \mathrm{cm}^3$
$N_A=6.022 \times 10^{23}$
Using the formula
$d=\frac{Z \times M}{N_A \times a^3}$
$M=\frac{d \times N_A \times a^3}{Z}$
$=\frac{10 \times 6.022 \times 10^{23} \times\left(300 \times 10^{-10}\right)^3}{4}$
$\Rightarrow \quad M=40.65 \mathrm{~g}$
So, 40.65 g contains $6.022 \times 10^{23}$ atoms
Let 4.5 g contains $x$ atom
$\Rightarrow \quad x=\frac{6.022 \times 10^{23}}{40.65} \times 4.5=6.6 \times 10^{22}$ atoms
Hence, the correct answer is option: 'D'.