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?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

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.

Approved by eNotes Editorial Team