The density of a pure substance 'A' whose atoms pack in cubic close pack arrangement is 1 gm/cc. If B atoms can occupy tetrahedral void and if all the tetrahedral voids are occupied by &#039

Question

The density of a pure substance 'A' whose atoms pack in cubic close pack arrangement is 1 gm/cc. If B atoms can occupy tetrahedral void and if all the tetrahedral voids are occupied by ' B ' atom. What is the density of resulting solid in gm/cc ? [Atomic mass ( A ) = 30 gm / mol & Atomic mass ( B ) = 50 gm / mol ]

TestHub · Accepted Answer

In CCP (FCC), the number of A atoms per unit cell is $4$ , and there are $8$ tetrahedral voids. If all voids are occupied, the ratio of B:A is $8 : 4 = 2 : 1$ .

The mass of the original A crystal is $4 \times 30 = 120$ g per unit cell.

The new mass is $120 + (8 \times 50) = 520$ g per unit cell.

The original density is $1 g / cc$ . Since the volume remains the same, the new density is:

$ρ = \frac{520}{120} \times 1 = 4.33 g / cc$

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