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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
The total amount of gold in \(1.0 脳 10^3\) gallons of seawater is approximately 3.878 grams.

Step by step solution

01

Convert the concentration of gold from ppt to ppb

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. \[ Concentration\ in\ ppb = 13\ ppt\ 脳 1000 \approx 13,000\ ppb\]
02

Convert the concentration of gold from ppb to grams per gallon

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: \[1.025\ g/mL\ 脳\ 3785.41\ mL \approx 3877.546\ g\] Now, we can find the number of grams of gold in a gallon of seawater: \[\frac{1\ g}{10^9\ g\ of\ seawater}\ 脳\ 3877.546\ g\ of\ seawater\ per\ gallon \approx 3.877546\ 脳\ 10^{-6}\ g/gallon\]
03

Calculate the amount of gold in the given volume of seawater

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parts Per Trillion (ppt)
Parts per trillion (ppt) is a unit used in chemistry to describe extremely dilute concentrations of substances. It represents one part of a substance per one trillion parts of another (often a solvent or mixture).
In the exercise, we are talking about ppt to express the small concentration of gold in seawater. When dealing with very small quantities, ppt provides a practical way to communicate how scarce a substance is in a solution.
  • 1 ppt is equivalent to 0.000000000001 (or 10-12) of the total solution.
  • In essence, it's like one drop of ink in a trillion drops of water!
Understanding ppt helps us grasp how small quantities can still be present in vast amounts of a medium, such as gold in the ocean. To make calculations easier, we often convert ppt to parts per billion (ppb), which is more manageable because it's 1000 times larger than ppt.
Density of Seawater
The density of seawater plays a crucial role in converting the amount of gold from parts to a mass measurement. Density is defined as the mass per unit volume and in the case of seawater, its average density is typically around 1.025 grams per milliliter (g/mL).
Seawater's density is higher than pure water due to its salt content, which adds to the overall mass. For practical calculations, especially when dealing with large volumes like gallons, knowing the density helps convert from volume-based units to mass-based units.
  • A gallon of water is approximately 3.78541 liters, or 3785.41 milliliters.
  • With a density of 1.025 g/mL, one gallon of seawater weighs approximately 3877.546 grams.
This conversion is essential in calculating the mass of gold because it allows converting the concentration (given in parts) into grams, the typical unit for mass measurements in chemistry.
Unit Conversion in Chemistry
Unit conversion in chemistry is the process of converting measurements from one unit to another to solve problems more easily. It often involves a series of steps where various conversion factors are applied.
The original exercise is a great example of unit conversion. The problem involves converting concentration units from parts per trillion (ppt) to parts per billion (ppb), then to grams per gallon, a more tangible unit for quantifying the mass of gold.
  • To convert ppt to ppb, multiply by 1000 because 1 ppt = 0.001 ppb.
  • Then to convert ppb to grams per gallon, use the density of seawater to find the gram equivalent.
  • Finally, multiply the grams per gallon concentration by the total gallon volume for the total mass of the solute.
These conversions are vital for chemists to ensure that they can measure and work with different scales appropriately, providing accurate results in experiments and real-world applications.

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Most popular questions from this chapter

Indicate the type of solute-solvent interaction (Section 11.2) that should be most important in each of the following solutions: \((\mathbf{a}) \mathrm{CCl}_{4}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),(\mathbf{b})\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water, \((\mathbf{c}) \mathrm{KBr}\) in water, (d) HCl in acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right).\)

(a) Does a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{KCl}\) have a higher freezing point, a lower freezing point, or the same freezing point as a \(0.10 \mathrm{~m}\) aqueous solution of urea \(\left(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\right)\), (b) The experimental freezing point of the KCl solution is higher than that calculated assuming that \(\mathrm{KCl}\) is completely dissociated in solution. Why is this the case?

(a) In Equation 13.1, which of the enthalpy terms for dissolving an ionic solid would correspond to the lattice energy? (b) Which energy term in this equation is always exothermic?

What is the osmotic pressure formed by dissolving \(50.0 \mathrm{mg}\) of acetylsalicylic acid \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) in \(0.100 \mathrm{~L}\) of water at \(37^{\circ} \mathrm{C} ?\)

(a) What is the mass percentage of iodine in a solution containing \(0.035 \mathrm{~mol} \mathrm{I}_{2}\) in \(125 \mathrm{~g}\) of \(\mathrm{CCl}_{4} ?\) (b) Seawater contains \(0.0079 \mathrm{~g}\) of \(\mathrm{Sr}^{2+}\) per kilogram of water. What is the concentration of \(\mathrm{Sr}^{2+}\) in ppm?
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