Question

A metal of density equal to $7.5\times {10}^3$ kg m${}^{-3}$ has an fcc crystal structure with lattice parameter a equal to $400$ pm. Calculate the number of unit cells present in $0.015$ kg of the metal.

• A. $6.250\times {10}^{22}$
• B. $3.125\times {10}^{23}$
• C. $3.125\times {10}^{22}$
• D. $1.563\times {10}^{22}$

Answer 1

Answer: (C)

$3.125\times {10}^{22}$

$density=\frac{z\times \frac{M}{1000}}{N_A\times (a\times {10}^{-12}{)}^3}$
$7.5\times {10}^3=\frac{4\times \frac{M}{1000}}{6\times {10}^{23}\times (400\times {10}^{-12}{)}^3}$
$M=\frac{7.5\times {10}^3\times 6\times {10}^{23}\times 64\times {10}^{-30}\times 1000}{4}$
$M=72\text{g/mol}$
In $0.015\text{kg}$ of metal, there are $\frac{0.015\times 1000}{72}\times 6\times {10}^{23}=1.25\times {10}^{23}$ atoms
And $\frac{1.25\times {10}^{23}}{4}$ = $3.125\times {10}^{22}$ unit cells.