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Chapter 13: Problem 96

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.

Short Answer

Expert verified
Mass_{gold} (g) ≈ 0.06 g Thus, there are approximately 0.06 grams of gold in \(1.0 \times 10^{3}\) gallons of seawater.

Step by step solution

01

Convert volume from gallons to liters

First, we need the volume in liters. 1 gallon is approximately 3.78541 liters. So, the conversion formula is: \[Volume (L) = Volume (gal) \times 3.78541\] Now, we can calculate the volume in liters: \[Volume (L) = 1.0 \times 10^3 \times 3.78541\]
02

Convert volume to mass using seawater density

Next, we need to convert the volume in liters to mass in grams using the density of seawater. The approximate density of seawater is 1.025 grams per milliliter (g/mL). As 1 liter contains 1000 milliliters, we can use the following formula: \[Mass (g) = Volume (L) \times 1000 \times Density (g/mL)\] Now, we can calculate the mass of seawater: \[Mass (g) = Volume (L) \times 1000 \times 1.025\]
03

Calculate the mass of gold using its concentration in seawater

We are given that the concentration of gold in seawater is 13 ppt (parts per trillion). This means that there are 13 grams of gold in 1 trillion grams of seawater. So, the mass of gold can be calculated as follows: \[Mass_{gold} (g) = Mass_{seawater} (g) \times \frac{13}{1 \times 10^{12}}\] Now, by putting the values from Step 1 and Step 2 in Step 3, we get:
04

Calculate the final answer

\[Mass_{gold} (g) = (1.0 \times 10^{3} \times 3.78541) \times (1000 \times 1.025) \times \frac{13}{1 \times 10^{12}}\] Solve the equation above to get the mass of gold in grams contained in \(1.0 \times 10^{3}\) gallons of seawater.

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