Problem 96 - Chemistry The Central Science

Question

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.

Step-by-Step Solution

Answer

There are approximately 0.1085 grams of gold present in \(1.0 \times 10^{3}\) gallons of seawater, assuming a concentration of 13 ppt.

Step 1: 1. Convert gallons to grams

First, we need to deal with the unit conversion, as the seawater is given in gallons, and we need to convert this to grams. There are 8.3454 kg (or 8,345.4 grams) in a gallon of seawater. So, we have: \[ 1.0 \times 10^{3} \text{ gallons} \times 8,345.4 \frac{\text{grams}}{\text{gallon}} = 8,345,400 \text{ grams} \] This means that there are 8,345,400 grams of seawater in 1,000 gallons.

Step 2: 2. Calculate the number of grams of gold present using the concentration

Now, we know that there are 13 grams of gold in a trillion (10^12) grams of seawater. So, to find out how many grams of gold are in 8,345,400 grams of seawater, we can set up the following proportion: \[ \frac{13 \text{ grams}}{10^{12} \text{ grams}} = \frac{x}{8,345,400 \text{ grams}} \] Here, x represents the number of grams of gold in 8,345,400 grams of seawater.

Step 3: 3. Solve for x

To solve for x, we can cross-multiply: \[ 13 \text{ grams} \times 8,345,400 \text{ grams} = 10^{12} \text{ grams} \times x \] Divide both sides by 10^12 grams: \[ x = \frac{13 \times 8,345,400}{10^{12}} \text{ grams} \] Calculate x: \[ x \approx 0.1085 \text{ grams} \] So, there are approximately 0.1085 grams of gold present in 1.0 x 10^3 gallons of seawater, assuming a concentration of 13 ppt.