Calculate the total number of tetrahedral and octahedral voids formed in 0.6 mol of a compound if it forms hcp structure.

Step-by-step solution

Number of atoms in 0.6 mol: Since 1 mol contains 6.022 10²³ atoms, the total number is 0.6 6.022 10²³ = 3.6132 10²³ . Voids in hcp structure: Octahedral voids equal the number of atoms, giving 3.6132 10²³ . Tetrahedral voids are twice the number of atoms, yielding 2 3.6132 10²³ = 7.2264 10²³ . Total voids: Summing both gives 3.6132 10²³ + 7.2264 10²³ = 10.8396 10²³ . Expressed with three significant figures, this is 1.084 10²⁴ . This matches option C.