MHT CET · Chemistry · Solid State
How many atoms of niobium are present in \(2.43 \mathrm{~g}\) if it forms bcc structure with density \(9 \mathrm{~g} \mathrm{~cm}^{-3}\) and volume of unit cell \(2.7 \times 10^{-23} \mathrm{~cm}^3 ?\)
- A \(3.01 \times 10^{23}\)
- B \(4.1 \times 10^{22}\)
- C \(5.0 \times 10^{22}\)
- D \(2.0 \times 10^{22}\)
Answer & Solution
Correct Answer
(D) \(2.0 \times 10^{22}\)
Step-by-step Solution
Detailed explanation
Let number of atoms are \(\mathrm{N}\)
Mass of one atom \(=2.43 / \mathrm{N}\)
\(\mathrm{d}=\frac{\mathrm{Z} \times \text { mass of one atom }}{\text { Volume of unit cell }}[\mathrm{Z}=2\) for \(\mathrm{BCC}]\)
\(
9=\frac{2 \times 2.43 / \mathrm{N}}{2.7 \times 10^{-23}} \quad \mathrm{~N}=\frac{2 \times 2.43}{2.7 \times 10^{-23} \times 9}=2 \times 10^{22}
\)
Mass of one atom \(=2.43 / \mathrm{N}\)
\(\mathrm{d}=\frac{\mathrm{Z} \times \text { mass of one atom }}{\text { Volume of unit cell }}[\mathrm{Z}=2\) for \(\mathrm{BCC}]\)
\(
9=\frac{2 \times 2.43 / \mathrm{N}}{2.7 \times 10^{-23}} \quad \mathrm{~N}=\frac{2 \times 2.43}{2.7 \times 10^{-23} \times 9}=2 \times 10^{22}
\)
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