Question

AN element of atomic mass 50 occurs in bcc structure with a cell edge of 300pm calculate the avagadro number if density is 6.1g/cm3

Answer 1

Answer:

Na= 1.6 x 10^23

Explanation:

Length of the edge , a = 300 pm =300 x 10-10 cm  

Volume of unit cell = ( 300 x 10-10 cm )3 = 90 x 10-24 cm3

Since it is bcc arrangement,  

Number of atoms in the unit cell, Z = 2  

Atomic mass of the element = 50  

Mass of the unit cell = Z x M/No  

d=z×M/NA×V

Na = z×M/d×V

Na = 2×50/(6.91×90x10-24)

Na= 1.6 x 10^23

Answer 2

answer : Avogadro's number = 6.07 × 10²³

explanation : atomic mass of the element, M = 50 g/mol = 50 × 10^-3 kg/mol

cell edge , a = 300pm = 3 × 10^-10 m

number of atoms per unit cell of bcc , Z = 2

density of element, d = 6.1 g/cm³ = 6.1 × 10³ kg/m³

using formula,

d = Z × M/NA × a³

⇒NA = ZM/da³

= (2 × 50 × 10^-3)/{6.1 × 10³ × (3 × 10^-10)³}

= 0.1/(6.1 × 10³ × 27 × 10^-30)

= 0.1/(6.1 × 27 × 10^-27)

= 0.1 × 10^27/(6.1 × 27)

= 1000/(6.1 × 27) × 10²³

= 6.07164542 × 10²³ ≈ 6.07 × 10²³

hence , Avogadro's number is 6.07 × 10²³