Question

An element crystallizes in fcc lattice with a cell edge of 400pm.calculate the density if 200g of this element contain 2.5 x 1024 atoms.

Answer 1

Answer: The density of element is $4.99g/cm^3$

Explanation: Density of the crystal lattice is given by the formula:

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

where,

Z = number of atoms in crystal lattice

M = Atomic mass

a = Edge length of the crystal

$\rho=Density\\N_A=\text{Avogadro's Number}=6.022\times 10^{23}$

According to the mole concept,

$6.022\times 10^{23}$ atoms are present in 1 mole of element.

We are given,

$2.5\times 10^{24}$ atoms are present in 200 g of element

So, $6.022\times 10^{23}$ atoms will be present in = $\frac{200}{2.5\times 10^{24}}\times 6.022\times 10^{23}=48.18g$ of element.

Z = 4 (For fcc)

a = 400 pm = $400\times 10^{-10}cm$

Putting values in above equation, we get:

$\rho=\frac{4\times 48.18g}{6.022\times 10^{23}\times (400\times 10^{-10}cm)^3}\\\\\rho=4.99g/cm^3$