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Most abundant element dissolved in sea water is chlorine at a concentration of \(19 \mathrm{~g} / \mathrm{kg}\) of sea water. The volume of earth's ocean is \(1.4 \times 10^{21} 1\). How many g-atoms of chlorine are potentially available from the oceans? Density of sea water is \(1 \mathrm{~g} / \mathrm{ml} .\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(7.5 \times 10^{20}\) (b) \(27 \times 10^{2}\) (c) \(27 \times 10^{24}\) (d) \(7.5 \times 10^{19}\)

Short Answer

Expert verified
The number of g-atoms of chlorine in the oceans is approximately \(27 \times 10^{24}\) g-atoms.

Step by step solution

01

Calculate the Mass of Chlorine in the Oceans

First, we need to find the total mass of chlorine in the oceans. Since the concentration of chlorine is given as 19 g/kg of seawater, and we are given the volume of Earth's ocean as well as the density of seawater, we can use these to calculate the mass. First, we convert the volume of the oceans from liters to kilograms using the density of seawater. Volume (in liters) multiplied by the density (1 g/ml) will give us the mass of seawater in grams, and we convert this to kilograms to match the concentration units. Then we multiply the total mass of seawater in kilograms by the chlorine concentration to get the mass of chlorine in grams.
02

Convert Mass of Chlorine to Moles

To find the amount of chlorine in g-atoms, we first need to convert the mass of chlorine from grams to moles. The molar mass of chlorine (Cl) is approximately 35.5 g/mol. We divide the total mass of chlorine by the molar mass to calculate the number of moles of chlorine.
03

Convert Moles to g-atoms

Now that we have the moles of chlorine, we can convert it to g-atoms using Avogadro's number, which is the number of g-atoms in a mole. Multiply the amount in moles by Avogadro's number (\(6 \times 10^{23}\) g-atoms/mol) to get the number of g-atoms of chlorine.

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Key Concepts

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

Chlorine Concentration Calculation
Understanding how to calculate the concentration of a substance such as chlorine in a solution is a fundamental skill in chemistry. Chlorine concentration typically refers to how much chlorine is present in a certain amount of solution. It is usually expressed in grams per liter (g/L), grams per kilogram (g/kg), or molarity (moles per liter).

In our exercise, the concentration of chlorine is given as 19 g/kg of sea water. To find the total mass of chlorine, we use the volume of Earth's oceans and the density of sea water. The density helps convert the volume from liters to grams, and subsequently to kilograms to align with the concentration units. We then multiply the mass of sea water by the chlorine concentration.

For example, if the ocean's volume is 1 liter (1,000 grams because the density of sea water is 1 g/mL), with a chlorine concentration of 19 g/kg, the mass of chlorine in that liter would be 0.019 grams. When dealing with the vast volume of Earth's oceans, this calculation provides insight into the immense quantity of chlorine present.
Mole Conversion in Chemistry
The concept of the mole is crucial in chemistry, acting as a bridge between the microscopic world of atoms and the macroscopic world we can measure. A mole corresponds to Avogadro's number of particles, which is approximately 6.02 x 10^23 particles/mol, representing a set number of molecules, atoms, or ions.

The molar mass connects the mass of a substance to its number of moles. To convert from grams to moles, we divide the mass by the molar mass. For chlorine, which has a molar mass of approximately 35.5 g/mol, if we have 19 grams of chlorine, we would divide 19 grams by 35.5 g/mol to find the number of moles.

When dealing with mole conversions, it's important to balance units correctly and understand that the molar mass is specific to each element or compound, based on atomic or molecular weights.
Avogadro's Number Application
Avogadro's number, often denoted as NA, is a defining constant in chemistry. It represents the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance.

The application of Avogadro's number is widespread, from determining the number of atoms in a given sample to stoichiometry calculations in chemical reactions. To convert moles to atoms (or molecules), we simply multiply the number of moles by Avogadro's number.

For example, if we have 1 mole of chlorine, we would have NA chlorine atoms, or 6.02 x 10^23 chlorine atoms. This concept enables us to visualize and calculate the amount of a substance on an atomic level, making it easier to understand and quantify reactions and compositions in chemistry.

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Most popular questions from this chapter

A quantity of \(0.25 \mathrm{~g}\) of a substance when vaporized displaced \(50 \mathrm{~cm}^{3}\) of air at \(0^{\circ} \mathrm{C}\) and 1 atm. The gram molecular mass of the substance will be (a) \(50 \mathrm{~g}\) (b) \(100 \mathrm{~g}\) (c) \(112 \mathrm{~g}\) (d) \(127.5 \mathrm{~g}\)

In a textile mill, a double-effect evaporator system concentrates weak liquor containing \(4 \%\) (by mass) caustic soda to produce a lye containing \(25 \%\) solids (by mass). What is the weight of water evaporated per \(100 \mathrm{~g}\) feed in the evaporator? (a) \(125.0 \mathrm{~g}\) (b) \(50.0 \mathrm{~g}\) (c) \(84.0 \mathrm{~g}\) (d) \(16.0 \mathrm{~g}\)

A mixture is made equal volume of \(\mathrm{CO}\) and air. A spark passed through so that all the oxygen is converted to carbon dioxide. What will be fractional decrease in the total volume of system assuming pressure and temperature remain constant? Air contains \(20 \%\) oxygen by volume. (a) \(0.1\) (b) \(0.2\) (c) \(0.15\) (d) \(0.3\)

A quantity of \(10 \mathrm{~g}\) of a hydrocarbon exactly requires \(40 \mathrm{~g}\) oxygen for complete combustion. The products formed are \(\mathrm{CO}_{2}\) and water. When \(\mathrm{CO}_{2}\) gas formed is absorbed completely in lime water, the mass of solution increases by \(27.5 \mathrm{~g}\). What is the mass of water formed in combustion? (a) \(22.5 \mathrm{~g}\) (b) \(27.5 \mathrm{~g}\) (c) \(50 \mathrm{~g}\) (d) \(10 \mathrm{~g}\)

The molecular formula of a compound is \(\mathrm{X}_{4} \mathrm{O}_{9} .\) If the compound contains \(40 \% \mathrm{X}\), by mass, what is the atomic mass of \(X\) ? (a) 24 (b) 12 (c) 26 (d) 13

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