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# The price of an atom of aluminum

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• The price of an atom of aluminum
• ### The price of an atom of aluminum

Cherrylauct > 01-15-2020, 01:07 AM

I have no idea, and I can\'t figure this out, and neither can my aunt and she\'s super smart. It\'s this project that\'s due on Monday, and if I can\'t get it turned in, then I get detention and a failing lab grade. The only information I have is it\'s \$2.96 for a roll of aluminum, it contains 2.96 square meters, and it weighs .995 grams...This can be done, as crazy as it sounds. Aluminum has a molar mass of 26.98g and the stated sample (the roll) is \$2.96 for 0.995g You would divide 0.955 by 26.98. This gives you 0.0369, which is the amount of moles of aluminum in the roll. Then you need to use Avogadro\'s number. This is 6.02*10^23, and is the amount of particles in a mole. Avogadro\'s number multiplied by 0.0369 gives you 2.22138*10^22. This is the amount of aluminum atoms in the roll. Now, you divide the price (\$2.96) by the amount of aluminum atoms (2.22138*10^22) to get the price per atom. Price per atom of aluminum in the stated sample: \$1.332505*10^-22 (This is equivalent to .\$0.0000000000000000000001332505) I hope this helps.3 square meters masses less than a gram - surely not But lets assume you have 0.995 grams. Dividing this number by 26.98(molar mass of AL) would give you moles of Al. Multiplying the number of moles X Avogadro\'s number (6.022 X 10^23) would give you atoms of Al. That number of atoms will cost you \$2.96 . Dividing the cost by the total number of atoms will give you dollars per atom.say it\'s a trick question, because it there is no denotational value for a single aluminum atom :/Other related question