Question

A lead cube that is 3.00cm on each side contains 8.91 * 10^23 atoms. What is the density of this cube in g/cm3?

Answer 1

ok first calculate volume, it says its cube so it is
V= (3 cm)^3 = 27 cm3
and now calculate mass:
n=N/L = 8,91*10^23 / 6,022*10^23 mol-1 = 1,479 mol ~ 1,5 mol
n=m/M -> m= n*M = 1,5 mol * 207,19 gmol-1 = 310,785 g
rho= m / V = 310,785g / 27 cm3 = 11,51 g/cm3