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The total volume of seawater is \(1.5 \times 10^{21} \mathrm{~L}\). Assume that seawater contains 3.1 percent sodium chloride by mass and that its density is \(1.03 \mathrm{~g} / \mathrm{mL}\). Calculate the total mass of sodium chloride in kilograms and in tons. \((1\) ton \(=2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{~g} .)\).

Short Answer

Expert verified
The total mass of sodium chloride in seawater is approximately \(4.8 \times 10^{19} \mathrm{~kg}\) or \(10.6 \times 10^{16} \mathrm{~tons}\).

Step by step solution

01

Calculate the Mass of Seawater

To find the total mass of the seawater, we need to use the formula: Mass = Density * Volume. From the exercise, we know that the volume of seawater is \(1.5 \times 10^{21} \mathrm{~L}\) and the density of seawater is \(1.03 \mathrm{~g}/\mathrm{mL}\). Note that 1 \mathrm{L} equals 1000 \mathrm{mL}, so we first need to convert the volume from liters to milliliters. The total mass is then calculated as follows: \((1.5 \times 10^{21} \mathrm{~L}) * (1000 \mathrm{~mL}/\mathrm{L}) * (1.03 \mathrm{~g}/\mathrm{mL}) = 1.545 \times 10^{24} \mathrm{~g}\).
02

Find the Mass of Sodium Chloride

Seawater is given to contain 3.1% sodium chloride by mass. So the mass of sodium chloride can be calculated by multiplying the total mass of seawater by the percentage of sodium chloride. Note that the percentage needs to be in decimal form, so 3.1% becomes 0.031. Therefore, the mass of sodium chloride is: \((1.545 \times 10^{24} \mathrm{~g}) * 0.031 = 4.8 \times 10^{22} \mathrm{~g}\).
03

Convert Mass to Kilograms

We have that 1 \mathrm{~kg} = 1000 \mathrm{~g}. So, to convert the mass of sodium chloride to kilograms, we simply divide the total mass in grams by 1000: \((4.8 \times 10^{22} \mathrm{~g}) / (1000 \mathrm{~g}/\mathrm{~kg}) = 4.8 \times 10^{19} \mathrm{~kg}\).
04

Convert Mass to Tons

Given that 1 ton equals 2000 pounds and 1 pound equals 453.6 grams, we can convert the mass into tons. First convert the mass of sodium chloride from grams to pounds, then from pounds to tons. Thus, \((4.8 \times 10^{22} \mathrm{~g}) / (453.6 \mathrm{~g}/\mathrm{~lb}) / (2000 \mathrm{~lb}/\mathrm{~ton}) = 10.6 \times 10^{16} \mathrm{tons}\).

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Key Concepts

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

Density Calculation
Understanding density is crucial when dealing with mass and volume conversions. Density is defined as the mass per unit volume of a substance, expressed in units like grams per milliliter (g/mL).
In the context of seawater, which has a density of 1.03 g/mL, it means that every milliliter of seawater weighs 1.03 grams.
To calculate the total mass of a large volume of seawater, you would use the formula:
- Mass = Density × Volume.
Given the seawater volume as 1.5 × 10^{21} liters, first convert this to milliliters (since 1 ext{L} = 1000 ext{mL}), and then multiply by the density to find the mass. Density is an important property as it helps in understanding how compact or heavy a substance is in a given volume, which is especially useful in fields like chemistry and physics.
Percentage Composition
Percentage composition gives details about how much of a substance is present in a mixture. In this exercise, seawater comprises 3.1% sodium chloride by mass.
This percentage is crucial because it tells us the exact proportion of sodium chloride relative to the total mass of seawater.
- To use this information, convert the percentage into decimal form, which is 0.031 in this case.
Multiplying this decimal by the total mass of the seawater gives the mass of sodium chloride present in it.
This concept of percentage composition is commonly used in chemistry to quantify the components in mixtures and solutions, enabling calculations for reactions and formulations.
Unit Conversion
Unit conversion is a process of converting one unit of measure into another. It is essential in scientific calculations to maintain consistency in measurements.
In the seawater exercise, several conversions take place.
- Liters to milliliters: since 1 liter equals 1000 milliliters, multiplying the given volume in liters by 1000 gives the volume in milliliters.
- Grams to kilograms: divide the mass in grams by 1000 to convert to kilograms, because 1 kilogram equals 1000 grams.
Unit conversions allow calculations to be accurately compared and compatible across different measures, significant in data analysis and experiments.
Ton to Kilogram Conversion
Ton to kilogram conversion is a specific type of unit conversion. Knowing this is critical when dealing with large masses, often expressed in tons, especially when calculating the weight of materials in bulk.
- 1 ton equals 2000 pounds. - 1 pound is equivalent to 453.6 grams.
To convert grams to tons, first convert grams to pounds by dividing the mass in grams by 453.6.
Then, convert the result from pounds to tons by dividing by 2000. This conversion is especially useful in industries and scientific calculations to understand large-scale weights and assist in logistical planning and resource management.

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Most popular questions from this chapter

A sheet of aluminum (Al) foil has a total area of \(1.000 \mathrm{ft}^{2}\) and a mass of \(3.636 \mathrm{~g}\). What is the thickness of the foil in millimeters? (Density of \(\left.\mathrm{Al}=2.699 \mathrm{~g} / \mathrm{cm}^{3} .\right)\)

The thin outer layer of Earth, called the crust, contains only 0.50 percent of Earth's total mass and yet is the source of almost all the elements (the atmosphere provides elements such as oxygen, nitrogen, and a few other gases). Silicon (Si) is the second most abundant element in Earth's crust (27.2 percent by mass). Calculate the mass of silicon in kilograms in Earth's crust. (The mass of Earth is \(5.9 \times 10^{21}\) tons. 1 ton \(=2000 \mathrm{lb}\); \(1 \mathrm{lb}=453.6 \mathrm{~g} .)\)

Carry out these operations as if they were calculations of experimental results, and express each answer in the correct units and with the correct number of significant figures: (a) \(5.6792 \mathrm{~m}+0.6 \mathrm{~m}+4.33 \mathrm{~m}\) (b) \(3.70 \mathrm{~g}-2.9133 \mathrm{~g}\) (c) \(4.51 \mathrm{~cm} \times 3.6666 \mathrm{~cm}\) (d) \(\left(3 \times 10^{4} \mathrm{~g}+6.827 \mathrm{~g}\right) /\left(0.043 \mathrm{~cm}^{3}-0.021 \mathrm{~cm}^{3}\right)\)

What is the number of significant figures in each of these measured quantities? (a) \(40.2 \mathrm{~g} / \mathrm{cm}^{3}\), (b) \(0.0000003 \mathrm{~cm}\) (c) \(70 \mathrm{~min}\) (d) \(4.6 \times 10^{19}\) atoms.

Express these numbers in scientific notation: (a) 0.000000027 , (b) 356 (c) 0.096 .

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