If seawater contains 3.2 × 10^{-9} g of gold per liter, how many grams of gold are present in 1.00 L of seawater?

• A. 1.0 × 10^{-3} g
• B. 6.02 × 10^{23} g
• C. 0.32 g
• D. 3.2 × 10^{-9} g

Answer: (D)

Verified step by step guidance

1. Identify the given information: the concentration of gold in seawater is 3.2 \times 10^{-9} grams per liter.
2. Recognize that the problem asks for the amount of gold in 1.00 liter of seawater, which means the volume is 1.00 L.
3. Use the relationship between concentration and amount: \n\nAmount of gold (g) = Concentration (g/L) \times Volume (L) \n\nExpressed as: \(\text{mass} = (3.2 \times 10^{-9} \text{ g/L}) \times (1.00 \text{ L})\)
4. Multiply the concentration by the volume to find the total grams of gold in 1.00 L of seawater. Since the volume is 1.00 L, the mass will be equal to the concentration value.
5. Confirm that the units of liters cancel out, leaving the mass in grams, which matches the given concentration value.