MHT CET · Chemistry · Solid State
Which is the volume of unit cell of a metal (at. mass \(25 \mathrm{~g} \mathrm{~mol}^{-1}\) ) having BCC structure and density \(3 \mathrm{~g} \mathrm{~cm}^{-3}\) ?
- A \(3.64 \times 10^{-23} \mathrm{~cm}^3\)
- B \(1.56 \times 10^{-24} \mathrm{~cm}^3\)
- C \(2.76 \times 10^{-23} \mathrm{~cm}^3\)
- D \(1.88 \times 10^{-24} \mathrm{~cm}^3\)
Answer & Solution
Correct Answer
(C) \(2.76 \times 10^{-23} \mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
\(\text {Density }=\frac{\text { mass of atomsin unitcell }}{\text {Volume of unitcell }} \)
\( \text {mass of atoms in unit cell }=\) \(\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{N}_{\mathrm{A}}}=\frac{2 \times 25}{6.023 \times 10^{23}} \)
\( \text {Volume of unit cell }=\) \(\frac{2 \times 25}{6.02 \times 10^{23} \times 3}=2.76 \times 10^{-23} \mathrm{~cm}^3\)
\( \text {mass of atoms in unit cell }=\) \(\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{N}_{\mathrm{A}}}=\frac{2 \times 25}{6.023 \times 10^{23}} \)
\( \text {Volume of unit cell }=\) \(\frac{2 \times 25}{6.02 \times 10^{23} \times 3}=2.76 \times 10^{-23} \mathrm{~cm}^3\)
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