MHT CET · Chemistry · Solid State
Calculate number of atoms per unit cell of an element having molar mass \(92.0 \mathrm{~g} \mathrm{~mol}^{-1}\) and density \(8.6 \mathrm{~g} \mathrm{~cm}^{-3}\) forming cubic unit cell structure. \(\left[\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}=21.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]\)
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(\text { Density }(\rho)=\frac{M \times n}{a^3 \times N_A}\)
\(\mathrm{n}=\frac{\rho \times \mathrm{a}^3 \times \mathrm{N}_A}{\mathrm{M}}=\) \(\frac{8.6 \mathrm{~g} \mathrm{~cm}^{-3} \times 21.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}}{92 \mathrm{~g} \mathrm{~mol}^{-1}}=2\)
\(\mathrm{n}=\frac{\rho \times \mathrm{a}^3 \times \mathrm{N}_A}{\mathrm{M}}=\) \(\frac{8.6 \mathrm{~g} \mathrm{~cm}^{-3} \times 21.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}}{92 \mathrm{~g} \mathrm{~mol}^{-1}}=2\)
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