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Questions 4-14 · Q10

Q.An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (1.16×10²⁴, 2.32×10²⁴)

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Step 1. The printed question gives the edge length (288 pm) and mass (200 g) but does not state the element's molar mass, which is needed to convert mass into atoms/unit cells -- this is a genuine gap in the available source text. Back-solving from the textbook's own printed answer shows a molar mass of about 52 g/mol (matching chromium, a well-known bcc metal) reproduces it exactly, so that value is used below.

Step 2. For bcc, n=2n=2; edge length a=288 pm=2.88×10−8 cma=288\text{ pm}=2.88\times10^{-8}\text{ cm}, so a3=2.389×10−23 cm3a^3=2.389\times10^{-23}\text{ cm}^3.

Step 3. Using M=52 g/molM=52\text{ g/mol}: density ρ=nMa3NA=2×522.389×10−23×6.022×1023≈7.23 g/cm3\rho=\dfrac{nM}{a^3N_A}=\dfrac{2\times52}{2.389\times10^{-23}\times6.022\times10^{23}}\approx7.23\text{ g/cm}^3. …

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