Question

An element X (molar mass = 60 g/mol) has density of 6.23 g cm-3. Identify the type of cubic nit cell, if the edge length of the unit cell is 4 x 10 -8 cm.

Answer 1

given info : density of an element X of molar mass 60g/mol is 6.23 g/cm³ , edge length of the unit cell is 4 × 10^-8 m.

To find : identify the unit cell of the given element X.

Solution : density , d = 6.23 g/cm³ = 6.23 × 10³ kg/m³

edge length of unit cell, a = 4 × 10^-8 m

so, volume of unit cell, V = a³ = (4 × 10^-8)³ = 64 × 10¯²⁴ m³

Avogadro's number, N = 6.023 × 10²³

Using formula, d = ZM/NV

⇒6.23 × 10³ = (Z × 60)/(6.022 × 10²³ × 64 × 10¯²⁴)

⇒6.23 × 10³ × 6.022 × 64 × 10¯¹ = 60Z

⇒4 = Z

Here no of atoms per unit cell = 4 , it means there are 6 atoms in faces of cube and 8 atoms in corners of cube.

therefore unit cell structure of given element is Face centered cubic lattice.