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Ch.18 - Chemistry of the Environment
Chapter 18, Problem 42

Gold is found in seawater at very low levels, about 0.05 ppb by mass. Assuming that gold is worth about $1300 per troy ounce, how many liters of seawater would you have to process to obtain $1,000,000 worth of gold? Assume the density of water is 1.03 g/mL and that your gold recovery process is 50% efficient.

Verified step by step guidance
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Step 1: Convert the value of gold from dollars to troy ounces. Since $1,000,000 worth of gold is needed and gold is valued at $1300 per troy ounce, calculate the number of troy ounces required by dividing $1,000,000 by $1300.
Step 2: Convert the number of troy ounces of gold to grams. Use the conversion factor that 1 troy ounce is approximately 31.1035 grams.
Step 3: Calculate the total mass of gold needed in grams, considering the efficiency of the recovery process. Since the process is 50% efficient, you will need twice the amount of gold calculated in Step 2.
Step 4: Determine the mass of gold present in seawater. Given that the concentration of gold in seawater is 0.05 ppb (parts per billion) by mass, express this concentration as grams of gold per gram of seawater.
Step 5: Calculate the volume of seawater needed. Use the density of seawater (1.03 g/mL) to convert the mass of seawater (from Step 4) to volume in liters, ensuring all units are consistent.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Parts per billion (ppb)

Parts per billion (ppb) is a unit of measurement used to describe the concentration of a substance in a solution. It indicates how many parts of a substance are present in one billion parts of the total solution. In this context, 0.05 ppb means that there are 0.05 grams of gold in one billion grams of seawater, which is crucial for calculating the total amount of seawater needed to extract a specific quantity of gold.
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Parts per Billion (ppb)

Density and volume conversion

Density is defined as mass per unit volume and is essential for converting between mass and volume. In this problem, the density of seawater is given as 1.03 g/mL, which allows us to convert the mass of seawater into volume. Understanding this relationship is vital for determining how many liters of seawater are required to yield the desired amount of gold.
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Density Conversion Example

Efficiency of recovery processes

The efficiency of a recovery process indicates the percentage of the target substance that can be successfully extracted from a mixture. In this scenario, a 50% recovery efficiency means that only half of the gold present in the seawater can be obtained. This concept is critical for accurately calculating the total amount of seawater needed to achieve the desired monetary value of gold after accounting for losses during extraction.
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Related Practice
Textbook Question

The enthalpy of evaporation of water is 40.67 kJ/mol. Sunlight striking Earth's surface supplies 168 W per square meter (1 W = 1 watt = 1 J/s). (b) The specific heat capacity of liquid water is 4.184 J/g°C. If the initial surface temperature of a 1.00 square meter patch of ocean is 26 °C, what is its final temperature after being in sunlight for 12 h, assuming no phase changes and assuming that sunlight penetrates uniformly to depth of 10.0 cm?

Textbook Question

The enthalpy of fusion of water is 6.01 kJ/mol. Sunlight striking Earth's surface supplies 168 W per square meter (1 W = 1 watt = 1 J/s). (b) The specific heat capacity of ice is 2.032 J/g°C. If the initial temperature of a 1.00 square emter patch of ice is -5.0°C, what is its final temperature after being in sunlight for 12 h, assuming no phase changes and assuming that sunlight penetration uniformly to a depth of 1.00 cm?

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Textbook Question

The Ogallala aquifer described in the Closer Look box in Section 18.3, provides 82% of the drinking water for the people who live in the region, although more than 75% of the water that is pumped from it is for irrigation. Irrigation withdrawals are approximately 18 billion gallons per day. (a) Assuming that 2% of the rainfall that falls on an area of 600,000 km2 recharges the aquifer, what average annual rainfall would be required to replace the water removed for irrigation?

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Textbook Question

The Ogallala aquifer described in the Closer Look box in Section 18.3, provides 82% of the drinking water for the people who live in the region, although more than 75% of the water that is pumped from it is for irrigation. Irrigation withdrawals are approximately 18 billion gallons per day. (b) What process or processes accounts for the presence of arsenic in well water?

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