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Chapter 1: Problem 55

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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Most popular questions from this chapter

A bank teller is asked to assemble "one-dollar" sets of coins for his clients. Each set is made of three quarters, one nickel, and two dimes. The masses of the coins are: quarter: \(5.645 \mathrm{~g}\); nickel: \(4.967 \mathrm{~g}\); dime: \(2.316 \mathrm{~g}\). What is the maximum number of sets that can be assembled from \(33.871 \mathrm{~kg}\) of quarters, \(10.432 \mathrm{~kg}\) of nickels, and \(7.990 \mathrm{~kg}\) of dimes? What is the total mass (in g) of this collection of coins?

A 1.0 -mL volume of seawater contains about \(4.0 \times\) \(10^{-12} \mathrm{~g}\) of gold. The total volume of ocean water is \(1.5 \times 10^{21} \mathrm{~L} .\) Calculate the total amount of gold in grams that is present in seawater and its worth in dollars, assuming that the price of gold is \(\$ 350\) an ounce. With so much gold out there, why hasn't someone become rich by mining gold from the ocean?

The surface area and average depth of the Pacific Ocean are \(1.8 \times 10^{8} \mathrm{~km}^{2}\) and \(3.9 \times 10^{3} \mathrm{~m},\) respectively. Calculate the volume of water in the ocean in liters.

How many seconds are in a solar year \((365.24\) days \() ?\)

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