Answer
(a) $6.02\times 10^{28}atoms/m^3$
(b) $3.28\times 10^{28}atoms/m^3$
Work Step by Step
(a) We can find the number density of aluminum as follows:
$n=\frac{M}{M_{mol}}$
$\implies n=(2700Kg)(\frac{1mol}{0.027Kg})=1.0\times 10^5mol$
We know that
$N=nN_A$
$\implies N=(1.0\times 10^{5}mol)(6.02\times 10^{23}atoms/mol)$
$\implies N=6.02\times 10^{28}atoms$
Now the the number density is given as
$\frac{N}{V}=\frac{6.02\times 10^{28}atoms}{1m^3}=6.02\times 10^{28}atoms/m^3$
(b) The number density of lead can be determined as follows:
$n=\frac{M}{M_{mol}}$
$\implies n=(11,300Kg)(\frac{1mol}{0.207Kg})$
Now the number density is given as
$\frac{N}{V}=\frac{nN_A}{V}$
We plug in the known values to obtain:
$\frac{N}{V}=(\frac{11,300Kg}{1m^3})(\frac{1mol}{0.207Kg})(\frac{6.02\times 10^{23}atoms}{1mol})=3.28\times 10^{28}atoms/m^3$
