Question

The density of lead (Pb) is 11.34 g/cm^3. How many atoms of Pb are present in a cube of Pb that has a length of 8.00 mm on each side? Please show all work and calculations in detail.

Answer 1

length of each side,

a = 8.00 mm = 0.800 cm

volume of cube,

V = a^3

= (0.800 cm)^3

= 0.512 cm^3

now use:

mass = density * volume

= 11.34 g/cm^3 * 0.512 cm^3

= 5.806 g

Molar mass of Pb = 207.2 g/mol

mass of Pb = 5.806 g

we have below equation to be used:

number of mol of Pb,

n = mass of Pb/molar mass of Pb

=(5.806 g)/(207.2 g/mol)

= 2.802*10^-2 mol

we have below equation to be used:

number of atoms = number of moles * Avogadro’s number

number of atoms = 2.802*10^-2 * 6.022*10^23 atoms

number of atoms = 1.687*10^22 atoms

Answer: 1.69*10^22 atoms