Sample Solution from
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Chapter 1.4
Problem 1.6BFP
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Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each mL of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square (m2) is the SI unit of the area. The English equivalent of the area is miles square (mi2). The conversion factor to convert the area from mi2 to ft2 is,

(0.3048)2 m2=1 ft2

Mass is a physical quantity and its SI unit is kilograms (kg). It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

(beginning unit)(Final unitbeginning unit)=Final unit

Expert Solution

Answer

The road map to calculate the mass of mercury in grams contained in each mL of lake water is as follows:

images

The mass of mercury in grams contained in each mL of lake water is 6.0×10−7 g/mL.

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each mL of lake water is as follows:

images

The mass of mercury discharged in the lake is 75,000 kg.

The conversion factor to convert the mass from kilograms (kg) to grams (g) is,

1 kg=1000 g

The mass of mercury in the lake in grams is,

Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g

The surface area of the lake is 4.5 mi2 and the depth of the lake is 35 ft.

The conversion factor to convert the area from mi2 to ft2 is,

1 mi2=(5280)2 ft2

The surface area of the lake in ft2 is calculated as:

Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2

The volume of  the lake is calculated as:

Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft)) (1)

Substitute 1.25452800×108 ft2 for the area of the lake and 35 ft for depth of lake in equation (1).

Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3

The conversion factor to convert volume from ft3 to m3 is,

1 ft3=0.02832​ m3

The conversion factor to convert volume from m3 to cm3 is,

1 m3=1×106 cm3

The conversion factor to convert volume from cm3 to mL is,

1 cm3=1​ mL

The volume of the lake in mL is,

Volume of lake(mL)=4.3908×109 ft3(0.02832​ m31 ft3)(1×106 cm31 m3)(1​ mL1 cm3)=1.24349×1014 mL

The mass of mercury in grams present in one mL of lake water is calculated as,

Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL) (2)

Substitute 7.5×107 g for the mass of mercury discharged in the lake and 1.24349×1014 mL for the volume of the lake in equation (2).

Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL

Conclusion

The mass of mercury in grams contained in each mL of lake water is 6.0×10−7 g/mL.

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