Question

A elements has a body centred cubic structure with a cell age of 288 pm the density of the element is 7.2 gram per centimetre cube how many atoms are present in 208 g of the element

Answer 1

We may write equations of number of particles and mass as the following:
$N=n\cdot N_A\\ m=n \cdot M\\ where\\ N= \text{Number of particles}\\ N_A=\text{Avogadro's number}\\ m=\text{Mass of substance}\\ M=\text{Molar mass of substance}\\ n=\text{Number of moles}\\ \text{Dividing the above two equations we get}\\ \frac{M}{N_A}=\frac{m}{N}\hfill-1\\ \text{For B.C.C lattice,}\\ z=\frac{1}{8}\cdot\text{Corner atoms}+\frac{1}{1}\cdot\text{Center atom}\\ =\frac{1}{8}\cdot8+1=2\\ \text{Now apply the formula}\\ \rho=\frac{zM}{a^3N_A}\\ \text{From equation 1}\\ \rho=\frac{zm}{a^3N}\\ \Rightarrow N=\frac{zm}{a^3\rho}\\ \text{Putting values, we get}\\ N=2.4187\times{10}^{24}$

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.