Question

The face-centered unit cell of nickel has an edge length of $352.39pm$. The density of nickel is $8.9gc{m}^{-3}$. Calculate the value of Avogadro's number. The atomic mass of nickel is $58.7$ and $1pm$ is equal to ${10}^{-10}cm$.

Answer 1

$a=352.39pm$ $\rho =8.9g/c{m}^3$, $M_A=58.7$, $N_A=?$ , $Z=4$ FCC

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

$8.9=\frac{4\times 58.7}{N_A\times 43.76\times {10}^{-24}}\Rightarrow N_A=6.0287\times {10}^{23}$