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Chapter 20: Problem 1

Find the mass in kilograms of \(7.50 \times 10^{24}\) atoms of arsenic, which has a molar mass of \(74.9 \mathrm{~g} / \mathrm{mol}\).

Short Answer

Expert verified
The mass is approximately 0.933 kg.

Step by step solution

01

Understand the Given Information

You are given the number of atoms of arsenic, which is \(7.50 \times 10^{24}\) atoms, and the molar mass of arsenic, which is 74.9 g/mol. The goal is to find the mass in kilograms.
02

Convert Atoms to Moles

Use Avogadro's number, \(6.022 \times 10^{23}\) atoms/mol, to convert the number of atoms to moles. Divide the given number of atoms by Avogadro's number: \( n = \frac{7.50 \times 10^{24}\text{ atoms}}{6.022 \times 10^{23}\text{ atoms/mol}} \approx 12.45 \text{ moles}\)
03

Calculate the Mass in Grams

Multiply the number of moles by the molar mass to get the mass in grams: \( m = 12.45\text{ moles} \times 74.9\text{ g/mol} \approx 933.405\text{ g} \)
04

Convert Grams to Kilograms

Since there are 1000 grams in a kilogram, divide the mass in grams by 1000: \( m = \frac{933.405\text{ g}}{1000} \approx 0.933\text{ kg}\)

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Key Concepts

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

Avogadro's number
Avogadro's number is a fundamental constant in chemistry. It tells us the number of atoms, molecules, or particles in one mole of a substance. The official value is approximately \( 6.022 \times 10^{23} \) atoms/mol.
This huge number helps us convert between the microscopic scale (atoms and molecules) and the macroscopic scale (grams and moles).
For example, if you want to know how many moles are in a given number of atoms, you can divide the number of atoms by Avogadro's number.
In the exercise, you were given \( 7.50 \times 10^{24} \) atoms of arsenic and used Avogadro's number to find the number of moles:
  • \( n = \frac{7.50 \times 10^{24}}{6.022 \times 10^{23}} \text{ atoms/mol} \ bsp \)
  • \( n \approx 12.45 \text{ moles} \ bsp \)
Using this value, we can proceed to other calculations.
Atomic Mass
The atomic mass of an element is the weighted average mass of an element's isotopes on Earth. It is usually measured in atomic mass units (amu) or grams per mole (g/mol). The atomic mass can be found on the periodic table and varies from one element to another.
In our exercise, the atomic mass of arsenic is given as \( 74.9 \text{ g/mol} \).
This means every mole of arsenic atoms weighs 74.9 grams.
To find the total mass in grams, we multiply the number of moles by the atomic mass:
  • \( mass \text{ in grams} = 12.45 \text{ moles} \times 74.9 \text{ g/mol} \ bsp \)
  • \( mass \text{ in grams} \approx 933.405 \text{ g} \ bsp \)
Now, we need to convert this mass into kilograms.
Unit Conversion
Unit conversion is an essential skill in chemistry and other sciences. It allows us to change measurements from one unit to another, making data easier to understand and work with.
In this exercise, we need to convert the mass from grams to kilograms since the final answer must be in kilograms and not in grams.
The conversion factor between grams and kilograms is simple:
  • \( 1 \text{ kilogram} = 1000 \text{ grams} \ bsp \)
To perform this conversion, divide the mass in grams by 1000:
  • \( mass \text{ in kilograms} = \frac{933.405 \text{ grams}}{1000} \ bsp \)
  • \( mass \text{ in kilograms} \approx 0.933 \text{ kg} \ bsp \)
With this conversion, you can see that \(7.50 \times 10^{24}\) atoms of arsenic have a mass of about 0.933 kilograms. It's that straightforward!

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Most popular questions from this chapter

When a molecule of a liquid approaches the surface, it experiences a force barrier that tries to keep it in the liquid. Thus it has to do work to escape and loses some of its kinetic energy when it leaves. (a) Assume that a water molecule can evaporate from the liquid if it hits the surface from the inside with a kinetic energy greater than the thermal energy corresponding to the temperature of boiling water, \(100^{\circ} \mathrm{C}\). Use this to estimate the numerical value of the work \(W\) required to remove a water molecule from the liquid. (b) Even though the average speed of a molecule in water below the boiling point corresponds to a kinetic energy less than \(W\), some molecules leave anyway and the water evaporates. Explain why this happens.

A container encloses two ideal gases. Two moles of the first gas are present, with molar mass \(M_{1} .\) The second gas has molar mass \(M_{2}=3 M_{1}\), and \(0.5 \mathrm{~mol}\) of this gas is present. What fraction of the total pressure on the container wall is attributable to the second gas? (The kinetic theory explanation of pressure leads to the experimentally discovered law of partial pressures for a mixture of gases that do not react chemically: The total pressure exerted by the mixture is equal to the sum of the pressures that the several gases would exert separately if each were to occupy the vessel alone.)

The best laboratory vacuum has a pressure of about \(1.00 \times 10^{-18} \mathrm{~atm}\), or \(1.01 \times 10^{-13} \mathrm{~Pa}\). How many gas molecules are there per cubic centimeter in such a vacuum at \(293 \mathrm{~K} ?\)

We know that for an adiabatic process \(P V^{\gamma}=\) a constant. Evaluate the constant for an adiabatic process involving exactly \(2.0 \mathrm{~mol}\) of an ideal gas passing through the state having exactly \(P=1.0 \mathrm{~atm}\) and \(T=300 \mathrm{~K}\). Assume a diatomic gas whose molecules have rotation but not oscillation.

A certain gas occupies a volume of \(4.3 \mathrm{~L}\) at a pressure of \(1.2 \mathrm{~atm}\) and a temperature of \(310 \mathrm{~K}\). It is compressed adiabatically to a volume of \(0.76 \mathrm{~L}\). Determine (a) the final pressure and (b) the final temperature, assuming the gas to be an ideal gas for which \(\gamma=1.4\).
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