MHT CET · Chemistry · Solid State
Calculate the volume of unit cell of element with density \(7.2 \mathrm{~g} \mathrm{~cm}^{-3}\) that forms bcc structure. (288 gram of this element \(3.35 \times 10^{24}\) atoms)
- A \(3.038 \times 10^{23} \mathrm{~cm}^3\)
- B \(4.18 \times 10^{23} \mathrm{~cm}^3\)
- C \(6.136 \times 10^{23} \mathrm{~cm}^3\)
- D \(3.912 \times 10^{23} \mathrm{~cm}^3\)
Answer & Solution
Correct Answer
(B) \(4.18 \times 10^{23} \mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
\(\mathrm{d}=\frac{\mathrm{Z}}{\mathrm{V}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\)
\(7.2=\frac{2}{V} \times \frac{288}{3.35 \times 10^{24}}\)
\(\mathrm{V}=4.18 \times 10^{22} \mathrm{~cm}^3\)
\(7.2=\frac{2}{V} \times \frac{288}{3.35 \times 10^{24}}\)
\(\mathrm{V}=4.18 \times 10^{22} \mathrm{~cm}^3\)
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