Question

An element has a cep lattice with a cell edge length of 300 sm. The
density of the element is log/em? How many atoms are present in
270 g of the element ?​

Answer 1

The actual question is: an element has a ccp lattice with a cell edge length of 300pm the density of the element is 10 g/cm³. How many atoms are present in 270g of element?

No of atoms in ccp lattice: 4 = Z

edge length, a = 300 pm = 300 × 10⁻¹⁰ cm

N = 6.02 × 10²³ = Avogadro's constant

density of element = 10 g/cm³

According to formula,

$d=\frac{Zm}{N a^{3} }$

No of atoms = no of moles * N

$\frac{270 * N}{m}...........(1) \\\\m = \frac{dNa^{3}}{Z} \\\\\\=inserting\ this\ in (1)\\\\= \frac{270*N* Z}{dNa^{3}}$

= 4 × 10²⁵ atoms