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Chapter 3: Problem 33

What is the mass, in kilograms, of an Avogadro's number of people, if the average mass of a person is \(160 \mathrm{lb}\) ? How does this compare with the mass of Earth, \(5.98 \times 10^{24} \mathrm{~kg}\) ?

Short Answer

Expert verified
The mass of an Avogadro's number of people, considering an average mass of 160 lb (72.57 kg), is approximately \(4.37 \times 10^{25} \mathrm{kg}\). This is about 7.31 times greater than the mass of Earth (\(5.98 \times 10^{24} \mathrm{kg}\)).

Step by step solution

01

Convert pounds to kilograms

To convert the average mass of a person from pounds to kilograms, we use the following conversion factor: 1 pound = 0.453592 kg So, the average mass of a person in kilograms is: \(160 \mathrm{lb} \times 0.453592 \frac{\mathrm{kg}}{\mathrm{lb}} \approx 72.57 \mathrm{kg}\)
02

Calculate the mass of Avogadro's number of people

Now that we have established the average mass of a person in kilograms, we can multiply it by Avogadro's number (6.022 x 10^23) to find the mass of an Avogadro's number of people. To do this, simply use the formula: Mass = (average mass in kg) x (Avogadro's number) Mass ≈ \(72.57 \mathrm{kg} \times 6.022 \times 10^{23} = 4.37 \times 10^{25} \mathrm{kg}\) So, the mass of Avogadro's number of people is approximately \(4.37 \times 10^{25} \mathrm{kg}\).
03

Compare with the mass of Earth

Now that we know the mass of Avogadro's number of people, we can compare it with the mass of Earth, which is given as \(5.98 \times 10^{24} \mathrm{kg}\). To find the ratio of the calculated mass to the mass of Earth, we can divide the mass of Avogadro's number of people by the mass of Earth: Ratio = \(\frac{4.37 \times 10^{25} \mathrm{kg}}{5.98 \times 10^{24} \mathrm{kg}} \approx 7.31\) This result means that Avogadro's number of people would have a mass approximately 7.31 times greater than the mass of Earth.

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Most popular questions from this chapter

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{3} \mathrm{O}\), molar mass $=62.0 \mathrm{~g} / \mathrm{mol}$ (b) empirical formula \(\mathrm{NH}_{2}\), molar mass $=32.0 \mathrm{~g} / \mathrm{mol}$

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