A metal M forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _________ × 10²¹ (Nearest integer).
(Given N_A=6.02×10²³)

Solution

Chemistry ma’am · 3 steps

Each atom in a close-packed structure gives one octahedral and two tetrahedral voids: three per atom.

1. Count the atoms. Number of atoms N = 0.02 × 6.02 × 10²³ = 1.204 × 10²².
2. Voids per atom. In any close-packed structure, hcp or ccp, there are N octahedral voids and 2N tetrahedral voids. So the total is 3N.

   3N = 3 × 1.204 × 10²² = 3.612 × 10²² = 36.12 × 10²¹

3. So. The total number of voids is 36.12 × 10²¹, which rounds to 36.

The common slip: 24. Counting only the tetrahedral voids gives 2N = 24.08 × 10²¹, so 24. The octahedral voids must be added too.

The idea to keep. For N atoms in hcp or ccp: N octahedral voids and 2N tetrahedral voids, so 3N in total. Find N from moles first, then multiply.

Answer from the NTA final answer key. Solution by PrepWiser.