MHT CET · Chemistry · Solid State
What is atomic mass of an element with \(\mathrm{BCC}\) structure and density \(10 \mathrm{~g} \mathrm{~cm}^{-3}\) having edge length \(300 \mathrm{pm}\) ?
- A \(51.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(60.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(81.3 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(96.8 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(96.8 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
For bcc unit cell, \(\mathrm{n}=2\).
\(\text {Density of bcc unit cell }=\rho=\frac{\mathrm{Mn}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \)
\( \mathrm{M}=\frac{\rho \times \mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}} \)
\( \mathrm{M}=\) \(\frac{10 \mathrm{~g} \mathrm{~cm}^{-3} \times\left(3 \times 10^{-8} \mathrm{~cm}\right)^3 \times 6.022 \times 10^{23} \mathrm{atom} \mathrm{mol}^{-1}}{2 \text { atoms }} \)
\( \mathrm{M}=81.3 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\text {Density of bcc unit cell }=\rho=\frac{\mathrm{Mn}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \)
\( \mathrm{M}=\frac{\rho \times \mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}} \)
\( \mathrm{M}=\) \(\frac{10 \mathrm{~g} \mathrm{~cm}^{-3} \times\left(3 \times 10^{-8} \mathrm{~cm}\right)^3 \times 6.022 \times 10^{23} \mathrm{atom} \mathrm{mol}^{-1}}{2 \text { atoms }} \)
\( \mathrm{M}=81.3 \mathrm{~g} \mathrm{~mol}^{-1}\)
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