... of aluminum atoms?(mole = 6.022x10^23)
Believe it or not, you have the proverbial not enough information in this problem. That's because you don't specify the thickness of the aluminum foil. You will certainly need to know that before you can determine the volume of Al and the number of atoms. You will also need the density of Al, but that's easy enough to come by. Just for grins, say the foil is 0.0020 cm thick, and the density of Al foil is 2.70 g/cm^3. If you have an Avogadro's number of Al atoms (1 mole), then you have 27.0 grams of Al (molar mass of Al). If the density of aluminum is 2.7 g/cm^3, then you have a piece of aluminum with a volume of 10.0 cm^3. V = L x W x H .... where H is the thickness of the foil You are given the width and have looked up a thickness, and have calculated the volume. So solve for length. L = V / (WH) = 10 cm^3 / (30.48 cm x 0.0020 cm) L = 164 cm You would need a section of foil 164 cm long to give you approximately 1 mol of aluminum atoms if the thickness is 0.02 mm.
Wonderful first answer, but I am going to try anyways. For 1 mole, you need 27 grams of aluminum. The density of aluminum is 2700 kg/m^3, which means that 2.7 grams is contained in 1 cm^3 We know, Volume = Length * Breadth * Thickness 10 = L * (30.48) * Thickness So, just convert the thickness that you have into cm, and then use it in the above equation. --edited from here-- The atomic radius is 125 pM So, you have each atom having 250 pM Total thickness = 250 * 10^-12 * 92000 m = 0.0023 cm Use this in the equation : 10 = L * 30.48 * 0.0023 L = 142 cm
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