Question:
An element crystallising in body centred cubic lattice has an edge length of 500 pm. If its density is 4 g cm\(^{-3}\), the atomic mass of the element (in g mol\(^{-1}\)) is (consider \( N_A = 6 \times 10^{23} \))
An element crystallising in body centred cubic lattice has an edge length of 500 pm. If its density is 4 g cm\(^{-3}\), the atomic mass of the element (in g mol\(^{-1}\)) is (consider \( N_A = 6 \times 10^{23} \))
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Always remember: BCC → $Z = 2$
Updated On: May 2, 2026
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The Correct Option is D
Solution and Explanation
Concept: Density of crystal lattice
\[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \] Where:
• $Z = 2$ (BCC)
• $a = 500$ pm = $5 \times 10^{-8}$ cm
Step 1: Convert edge length
\[ a = 500 \text{ pm} = 5 \times 10^{-8} \text{ cm} \]
Step 2: Substitute values
\[ 4 = \frac{2M}{(5 \times 10^{-8})^3 \times 6 \times 10^{23}} \]
Step 3: Solve denominator
\[ (5 \times 10^{-8})^3 = 125 \times 10^{-24} = 1.25 \times 10^{-22} \] \[ 1.25 \times 10^{-22} \times 6 \times 10^{23} = 7.5 \times 10^1 = 75 \]
Step 4: Solve for M
\[ 4 = \frac{2M}{75} \Rightarrow 2M = 300 \Rightarrow M = 150 \] \[ 4 = \frac{2M}{75} \Rightarrow M = \frac{4 \times 75}{2} = 150 \] Final Answer: \[ \boxed{150} \]
\[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \] Where:
• $Z = 2$ (BCC)
• $a = 500$ pm = $5 \times 10^{-8}$ cm
Step 1: Convert edge length
\[ a = 500 \text{ pm} = 5 \times 10^{-8} \text{ cm} \]
Step 2: Substitute values
\[ 4 = \frac{2M}{(5 \times 10^{-8})^3 \times 6 \times 10^{23}} \]
Step 3: Solve denominator
\[ (5 \times 10^{-8})^3 = 125 \times 10^{-24} = 1.25 \times 10^{-22} \] \[ 1.25 \times 10^{-22} \times 6 \times 10^{23} = 7.5 \times 10^1 = 75 \]
Step 4: Solve for M
\[ 4 = \frac{2M}{75} \Rightarrow 2M = 300 \Rightarrow M = 150 \] \[ 4 = \frac{2M}{75} \Rightarrow M = \frac{4 \times 75}{2} = 150 \] Final Answer: \[ \boxed{150} \]
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