Problem 80 - Chemistry: The Central Science

Question

A package of aluminum foil contains \(50 \mathrm{ft}^{2}\) of foil, which weighs approximately \(8.0\) oz. Aluminum has a density of \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\). What is the approximate thickness of the foil in millimeters?

Step-by-Step Solution

Answer

The approximate thickness of the aluminum foil is 0.0181 millimeters.

Step 1: Convert weight to grams and area to square centimeters

First, we need to convert the given weight in ounces to grams and the area of aluminum foil in square feet to square centimeters. We use the conversion factors 1 ounce = 28.35 grams, and 1 ft² = 929.03 cm². Given: Weight = 8.0 oz Area = 50 ft² Weight in grams: \(Weight(g) = 8.0\ oz \times 28.35\frac{g}{oz} = 226.8\ g\) Area in square centimeters: \(Area(cm^2) = 50\ ft^2 \times 929.03\frac{cm^2}{ft^2} = 46451.5\ cm^2\)

Step 2: Calculate volume using weight and density

To find the volume of the aluminum foil, we can use the formula relating weight, density and volume: \(Volume = \frac{Weight}{Density}\) Given: Weight = 226.8 g Density of aluminum = 2.70 g/cm³ Volume: \(Volume\ (cm^3) = \frac{226.8\ g}{2.70\frac{g}{cm^3}} = 84\ cm^3\)

Step 3: Calculate the thickness of the aluminum foil

Now that we have found the volume of the aluminum foil, we can find the thickness using the formula relating volume, area, and thickness: \(Volume = Area \times Thickness\) Given: Volume = 84 cm³ Area = 46451.5 cm² Isolating the thickness, we get: \(Thickness = \frac{Volume}{Area}\) Thickness: \(Thickness\ (cm) = \frac{84\ cm^3}{46451.5\ cm^2} = 0.00181\ cm\)

Step 4: Convert thickness to millimeters

Finally, to express the thickness in millimeters, we convert it from centimeters to millimeters using the conversion factor 1 cm = 10 mm. Thickness in millimeters: \(Thickness\ (mm) = 0.00181\ cm \times 10\frac{mm}{cm} = 0.0181\ mm\) So, the approximate thickness of the aluminum foil is 0.0181 millimeters.