Question

An element crystallizes as face centred cubic lattice with density $5.20g/c{m}^3$ and edge length of the side of unit cell as $300pm$. Calculate the mass of the element which contains $3.01\times {10}^{24}$ atoms?

• A. 210
• B. 105
• C. 55
• D. 455

Answer 1

The correct option is B 105

Density $\rho =\frac{Z\times M}{N_0\times a^3}$
$\rho =5.20g/c{m}^3=$ density
$Z=4atoms=$ numeber of atoms per unit cell.
$M=$ molecular weight of element

$N_0=6.023\times {10}^{23}atoms/mol=$Avogadro's number

$a=300pm=300\times {10}^{-10}cm=$ edge length

Density $\rho =\frac{Z\times M}{N_0\times a^3}$

Density $5.20g/c{m}^3=\frac{4atoms\times M}{6.023\times {10}^{23}atoms/mol\times (300\times {10}^{-10}cm{)}^3}$

$M=21.14g/mol$

The molecular weight of element is 21.14 g/mol.

1 mole $(6.023\times {10}^{23}\text{atoms})$ of the element weighs 21.14 g.
1 atom of element will weigh $\frac{21.14g/mol}{6.023\times {10}^{23}atoms/mol}$ grams.
$3.01\times {10}^{24}$ atoms of element will weigh $\frac{21.14g/mol}{6.023\times {10}^{23}atoms/mol}\times 3.01\times {10}^{24}=105$ grams.