Question:
Aluminium (Atomic mass = 27) crystallizes in a cubic system with edge length of 4 \AA. Its density is 2.7 g cm$^{-3}$. The number of aluminium atoms present per unit cell is}
Aluminium (Atomic mass = 27) crystallizes in a cubic system with edge length of 4 \AA. Its density is 2.7 g cm$^{-3}$. The number of aluminium atoms present per unit cell is}
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FCC structure has \(Z = 4\).
Updated On: May 8, 2026
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The Correct Option is C
Solution and Explanation
Concept:
\[
\rho = \frac{Z \cdot M}{a^3 \cdot N_A}
\]
Step 1: Convert units. \[ a = 4\AA = 4 \times 10^{-8}\,cm \]
Step 2: Substitute values. \[ 2.7 = \frac{Z \cdot 27}{(4\times10^{-8})^3 \cdot 6.022\times10^{23}} \]
Step 3: Simplify denominator. \[ a^3 = 64 \times 10^{-24} \] \[ 2.7 = \frac{27Z}{64\times10^{-24}\times6.022\times10^{23}} \]
Step 4: Compute. \[ 2.7 = \frac{27Z}{3.854} \] \[ Z = \frac{2.7 \times 3.854}{27} \approx 4 \]
Step 5: Final answer. \[ \boxed{4} \]
Step 1: Convert units. \[ a = 4\AA = 4 \times 10^{-8}\,cm \]
Step 2: Substitute values. \[ 2.7 = \frac{Z \cdot 27}{(4\times10^{-8})^3 \cdot 6.022\times10^{23}} \]
Step 3: Simplify denominator. \[ a^3 = 64 \times 10^{-24} \] \[ 2.7 = \frac{27Z}{64\times10^{-24}\times6.022\times10^{23}} \]
Step 4: Compute. \[ 2.7 = \frac{27Z}{3.854} \] \[ Z = \frac{2.7 \times 3.854}{27} \approx 4 \]
Step 5: Final answer. \[ \boxed{4} \]
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