The aluminum foil in a package containing 75 square feet of foil weighs 12 ounces. aluminum has a density of 2.70 g / cm3. what is the approximate thickness of the foil in mm?

Answer;

= 0.018 mm

Explanation;

Mass = 12 ounces

1 ounce = 28.4 g

thus mass in g = 12 * 28.4 = 340.8 g

Volume = mass/density

= 340.8 g / 2.70 g/cm³

= 126 cm³

The thickness from volume and area;

thickness = volume / area

Area = 75 square ft

1 Square ft = 929.03 square centimeter

Thickness = 126 cm³ / (75 * 929.03)

= 0.001808 cm

but, 1 cm = 10 mm

Therefore; 0.001808 * 10 = 0.01808 mm

= 0.018 mm (3 decimal places)

Solution:

Since we know that 1 ounce=28.4g/oz

So for 12 oz X 28.4 g/oz = 340.8 g

And,

340.8 g X 1cm3/2.7g = 126.22 cm3

Again,

75 ft2 X 12 in/ft X 12 in/ft X 2.54 cm/in X 2.54 cm/in = 69,677.28 cm2

(Here ft^2=ft*ft)

Now,

69,677.28 cm2 X Thickness = 126.22 cm3

Finally Solve for Thickness: 126.22 cm3 / 69,677.28 cm2 = 0.00181 cm

Thus for mm we have to:

0.00181 cm X 10 mm/1 cm = 0.0181 mm.