KCET · Chemistry · Solid State
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 \(6.6 \times 10^{19}\)
- D \(6.6 \times 10^{22}\)
Answer & Solution
Correct Answer
(D) \(6.6 \times 10^{22}\)
Step-by-step Solution
Detailed explanation
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
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
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