The total mass of sodium chloride in the given sample of sea water in kilograms and in tons is to be determined. Concept introduction: In order to convert from one unit to another, there are certain relationships between those units. These relationships are called conversion factors. Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors. The appropriate conversion factor for any equality is selected in such a way that it results in the proper unit cancellation.
The total mass of sodium chloride in the given sample of sea water in kilograms and in tons is to be determined. Concept introduction: In order to convert from one unit to another, there are certain relationships between those units. These relationships are called conversion factors. Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors. The appropriate conversion factor for any equality is selected in such a way that it results in the proper unit cancellation.
The total mass of sodium chloride in the given sample of sea water in kilograms and in tons is to be determined.
Concept introduction:
In order to convert from one unit to another, there are certain relationships between those units. These relationships are called conversion factors.
Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors.
The appropriate conversion factor for any equality is selected in such a way that it results in the proper unit cancellation.
Expert Solution & Answer
Answer to Problem 87AP
Solution:4.8×1019 kg and 5.3×1016 tons
Explanation of Solution
Given information:
The volume of seawater is 1.5×1021 L.
The seawater contains 3.1% sodium chloride by mass.
1 ton=2000 lb
1 lb=453.6 g.
The total amount of sodium chloride present in 1.5×1021 L seawater can be evaluated as follows:
Density=massvolume(1.03g1 mL)=Mass of NaCl1.5×1021L(1.03g1 mL)=Mass of NaCl1.5×1021L
Mass of NaCl=(1.03g1 mL)×(1.5×1021 L×(1000 mL1 L)3.1% mass of NaCl=(1.03g1 mL)×(1.5×1021 L)×(1000 mL1 L)×(3.1 g NaCl100 g)
Amount of NaCl=(1.03 g1.0mL)(1.5×1021L)(1000mL1L)(3.1 gNaCl100g)=4.8×1022 g
Hence, the total amount of sodium chloride present in the seawater is 4.8×1022 g.
Convert grams into kilograms by using the following conversion factor:
Mass of NaCl=(4.8×1022g)(1 kg1000g)=4.8×1019 kg
Hence, the mass of sodium chloride in kilograms is 4.8×1019 kg.
One pound is equivalent to 453.6 g.
One ton is equivalent to 2000 lb.
Convert the mass of sodium chloride in tons as follows:
Mass of NaCl=(4.8×1022g)(1lb453.6 g)(1 ton2000lb)=5.3×1016 tons
Conclusion
The total mass of sodium chloride in kilograms and in tonsin the given sample of sea water is calculated to be 4.8×1019 kg and 5.3×1016 tons, respectively.
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