Questions: Gold can be hammered into extremely thin sheets called gold leaf. An architect wants to cover a 100 ft x 81 ft ceiling with gold leaf that is five-millionths of an inch thick. The density of gold is 19.32 g / cm^3. If gold costs 1768 per troy ounce (1 troy ounce = 31.10348 g), how much will it cost the architect to buy the necessary gold? Express your answer in dollars to one significant figure.

Solution Steps

First, we need to calculate the volume of the gold leaf required to cover the ceiling. The dimensions of the ceiling are $100 \, \text{ft} \times 81 \, \text{ft}$, and the thickness of the gold leaf is $5 \times 10^{-6} \, \text{inches}$.

Convert the dimensions to inches:

• $100 \, \text{ft} = 100 \times 12 = 1200 \, \text{inches}$
• $81 \, \text{ft} = 81 \times 12 = 972 \, \text{inches}$

Calculate the volume in cubic inches: $$\text{Volume} = 1200 \, \text{in} \times 972 \, \text{in} \times 5 \times 10^{-6} \, \text{in} = 5.832 \, \text{in}^3$$

Convert the volume from cubic inches to cubic centimeters using the conversion $1 \, \text{in}^3 = 16.387 \, \text{cm}^3$: $$\text{Volume} = 5.832 \, \text{in}^3 \times 16.387 \, \text{cm}^3/\text{in}^3 = 95.57 \, \text{cm}^3$$

Using the density of gold, $19.32 \, \text{g/cm}^3$, calculate the mass of the gold: $$\text{Mass} = 95.57 \, \text{cm}^3 \times 19.32 \, \text{g/cm}^3 = 1845.2 \, \text{g}$$

Convert the mass from grams to troy ounces using the conversion $1 \, \text{troy ounce} = 31.10348 \, \text{g}$: $$\text{Mass} = \frac{1845.2 \, \text{g}}{31.10348 \, \text{g/troy ounce}} = 59.31 \, \text{troy ounces}$$

Calculate the cost of the gold using the price of $\$1768$ per troy ounce: $$\text{Cost} = 59.31 \, \text{troy ounces} \times 1768 \, \text{\$/troy ounce} = 104,800.08 \, \text{\$}$$

Final Answer