Question

An element has bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm3. How many atoms are present in 208 g of the element?

Answer 1

Answer:

M = 24.18 × 10^23 g/mol

Explanation:

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Answer 2

${\huge{\underline{\underline{\mathcal{\red{♡ANSWER♡}}}}}}$.

• ❤.. ${Volume of the unit cell}$=${(288pm)^3}$
• ❤..=${(288×10^-12 m)^-3 = (288×10^-10 cm)^3}$
• ❤..=${2.39×10^-23 cm^3}$
• ❤..${Volume of208 g of element}$
• ❤..$\frac{mass}{density}$=$\frac{208g}{7.2gcm^-3}$=${28.88cm^3}$
• ❤..${Number of unit cells in this volume}$

=$\frac{28.88cm^3}{2.39×10^-23Cm^3/unit cell}$=${12.08×10^23 unit cells}$

• ⭐.. Since each${bcc}$cubic unit cell contains ${2}$ atoms,therefore,the total number of atoms in ${208g = 2}$${(atoms/unit cell)×12.08×10^23}$ unit cells=${24.16×10^23}$ atoms.