MHT CET · Chemistry · Solid State
Calculate the volume of unit cell if an element having molar mass \(180 \mathrm{~g} \mathrm{~mol}^{-1}\) forms fcc unit cell. \(\left[\rho \cdot \mathrm{N}_{\mathrm{A}}=120 \times 10^{21} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}\right]\)
- A \(6.00 \times 10^{-21} \mathrm{~cm}^3\)
- B \(5.00 \times 10^{-21} \mathrm{~cm}^3\)
- C \(4.00 \times 10^{-21} \mathrm{~cm}^3\)
- D \(7.00 \times 10^{-21} \mathrm{~cm}^3\)
Answer & Solution
Correct Answer
(A) \(6.00 \times 10^{-21} \mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
Volume of unit cell \(=\mathrm{a}^3=\) ?
Density \((\rho)=\frac{M \times n}{a^3 \times N_A}\)
\(\mathrm{a}^3=\frac{\mathrm{M} \times \mathrm{n}}{\rho \times \mathrm{N}_{\mathrm{A}}}=\frac{180 \times 4}{120 \times 10^{21}}=6.00 \times 10^{-21} \mathrm{~cm}^3\)
Density \((\rho)=\frac{M \times n}{a^3 \times N_A}\)
\(\mathrm{a}^3=\frac{\mathrm{M} \times \mathrm{n}}{\rho \times \mathrm{N}_{\mathrm{A}}}=\frac{180 \times 4}{120 \times 10^{21}}=6.00 \times 10^{-21} \mathrm{~cm}^3\)
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