# The mineral galena is composed of lead (II) sulfide and has an average density of 7.46 g/cm^3......How many molecules of lead (II) sulfide are in 1.00 ft.^3 of galena?

Given

density of mineral galena = 7.46 gm/cm^3

Volume = 1.00 ft^3 = 28316.8466 cm^3.

From density we know that 1 cm^3 contain 7.46 gm of galena.

So for 28316.8466 cm^3 of volume how much galena is present we have to find...

= [(28316.8466 cm^3)(7.46)]/[1 cm^3]

= 211243.676 gm of galena.

Therefore 211243.676 gm of galena is present in 1.00 ft^3.

Next step is finding moles for 211243.676 gm of galena.

Moles = mass/molar mass

Moles = 211243.676/239.3 [Molar mass of lead (II) sulfide = 239.3

Moles = 882.75669 of lead (II) sulfide.

From the law of avogadro number we know

1 mole of substance consist of = 6.022 X 10^23 molecules.

So for 882.75669 moles of lead (II) sulfide how many molecules of lead (II) sulfide is present we have to find...

Number of molecules of lead (II) sulfide

= 882.75669 * 6.022 X 10^23/1

= 5.31596079 × 10^26 molecules.

**5.31596079 × 10^26 molecules of lead (II) sulfide is present in 1.00 ft.^3 of galena.**

Now we know that

**1.00 ft.^3 = 28316.8466 cm^3**

So 7.46 g/cm^3 woule be

(7.46 g/ 1 cm^3) * (28316.8466 cm^3/1.00 ft.^3)

211243.676 g/ft^3

Now

**Density = mass/volume**

Mass = Density * volume

So the mass of lead (II) sulfide in 1 ft^3 would be

Mass of lead (II) sulfide = 211243.676 g/ft^3 * 1 ft^3 =211243.676 g lead (II) sulfide

We know that the molar mass of lead (II) sulfide is 239.28 g/mol

and 1 moles of lead (II) sulfide = 6.023*10^23 lead (II) sulfide molecule

so,

211243.676 g PbS * (1 mol PbS/239.28 g PbS)*(6.023*10^23 molecule PbS/ 1 mol PbS)

**5.32*10^26 molecule of PbS**