An atomic substance A of molar mass 12 g mol – 1 has a cubic crystal structure with edge length of 300 pm . The no. of atoms present in one unit cell of A is (Nearest integer) Given the density of A is 3 . 0 g m m - 1 and NA A = 6 . 02 × 10 23 mol - 1

Step-by-step solution

The density of crystal can be calculated as follows, ρ = Z × M N A × a 3 ρ = density of the unit cell Z = Effective atomic number M = Molar mass N A = Avogadro's number a = Edge length of the unit cell 3 = Z × 12 6 . 02 × 10 23 × ( 3 ) 3 × 10 - 24 Z = 6 . 02 × 10 - 1 × 27 4 = 4