The concentration of gold in seawater has been reported to be between 5 ppt (parts per trillion) and 50 ppt. Assuming that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in \(1.0 \times 10^{3}\) gal of seawater.

The concentration of gold in seawater is given in parts per trillion (ppt), but to make calculations easier, we will convert it to parts per billion (ppb). To do this, we will multiply the concentration by 1000 since 1 ppt = 0.001 ppb. \[ Concentration\ in\ ppb = 13\ ppt\ × 1000 \approx 13,000\ ppb\]

We have the concentration of gold in ppb; now, we need to convert it to grams per gallon. First, let's understand the units: 1 ppb (parts per billion) = 1 g of gold in \(10^9\) g of seawater. We're given the seawater volume in gallons, so we also need to know the density of seawater to convert grams to gallons. The density of seawater is approximately 1.025 g/mL. Using this, we can convert grams to gallons: 1 gallon = 3.78541 liters = 3785.41 mL (since 1 L = 1000 mL). Therefore, 1 gallon of seawater weighs: \[1.025\ g/mL\ ×\ 3785.41\ mL \approx 3877.546\ g\] Now, we can find the number of grams of gold in a gallon of seawater: \[\frac{1\ g}{10^9\ g\ of\ seawater}\ ×\ 3877.546\ g\ of\ seawater\ per\ gallon \approx 3.877546\ ×\ 10^{-6}\ g/gallon\]

We have the concentration of gold in grams per gallon, and the total volume of seawater is given as \(1.0 × 10^3\) gallons. Now, we can calculate the total amount of gold in grams: \[Amount\ of\ gold\ = \ Concentration\ in\ g/gallon\ × Volume\ in\ gallons\] \[Amount\ of\ gold\ = 3.877546\ ×\ 10^{-6}\ g/gallon\ ×\ 1.0\ ×\ 10^3\ gallons\] \[Amount\ of\ gold\ ≈ 3.877546\ g\] Therefore, the total amount of gold in \(1.0 × 10^3\) gallons of seawater is approximately 3.878 grams.