Question

If at room temperature crystal structure of Ca atom is fcc, then number of unit cells present in 20g of Ca is?​

Answer 1

given, weight of Ca = 20g

atomic weight of Ca = 40 g/mol

so, number of mole of Ca = given weight/atomic weight

= 20/40 = 0.5 mol

so, number of atoms of Ca = 0.5 mol × Avogadro's number

= 0.5 × 6.023 × 10²³

= 3.115 × 10²³

a/c to question,

If at room temperature crystal structure of Ca atom is FCC.

we know, number of atoms per unit cell of fcc = 4

then, total number of unit cells in 3.115 × 10²³ atoms = 3.115 × 10²³/4

= 0.77875 × 10²³

= 7.7875 × 10²²

hence, number of unit cells = 7.7875 × 10²²