Question

The density of a pure substance A whose atoms are in cubic closed packed arrangement is $1$ g/cc. If all the tetrahedral voids are occupied by B atoms, what is the density (in g cm${}^{-3}$) of resulting solid?[Atomic mass of $A=30$ g/mol and atomic mass of $B=50$ g/mol]

• A. $3.33$
• B. $4.33$
• C. $2.33$
• D. $5.33$

Answer 1

Answer: (B)

$4.33$

The formula for density is given below:

$d=\frac{nM}{V{N}_A}$ and therefore, $V=\frac{4\times 30}{1\times N_A}$

After adding B,

$d=\frac{(4\times 30+8\times 50)}{V{N}_A}=\frac{520}{120}=4.33$ (CCP has $8$ tetrahedral voids.)

Therefore, the density is $4.33$ g cm${}^{-3}$.

Option B is correct.