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?
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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)
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
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
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.
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