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Chapter 18: Problem 19

How many moles are in a \(1.00 \mathrm{~kg}\) bottle of water? How many molecules? The molar mass of water is \(18.0 \mathrm{~g} / \mathrm{mol} .\)

Short Answer

Expert verified
The number of moles in a \(1.00 \mathrm{~kg}\) bottle of water is approximately \(55.56\) moles, and the number of molecules is approximately \(3.34 \times 10^{25}\) molecules.

Step by step solution

01

Conversion

First, you need to convert the mass from kilograms to grams. So, \(1.00 \mathrm{~kg}\) of water equals \(1000 \mathrm{~g}\) of water.
02

Calculation of moles

Next, to calculate the number of moles, you have to divide the mass of substance by its molar mass. Since the molar mass of water is \(18.0 \mathrm{~g/mol}\) , the number of moles is \(\frac{1000}{18.0}\)
03

Number of Molecules

After you have the number of moles, you can find the total number of molecules. The number of entities in one mole of any substance is known as Avogadro鈥檚 number, approximately \(6.02214076 \times 10^{23}\). So the number of molecules will be the number of moles multiplied by Avogadro鈥檚 number.

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Key Concepts

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

Molar Mass
When dealing with chemical substances, the molar mass is a key concept. It provides the link between the mass of a substance and the amount of substance in moles. Molar mass is expressed in units of grams per mole (g/mol). For example, water (H\(_2\)O) has a molar mass of 18.0 g/mol.
To find how many moles a given mass of a substance contains, you simply divide the mass of the substance by its molar mass.
  • Molar mass allows us to convert between mass and number of moles.
  • This is immensely helpful in stoichiometric calculations, where balancing reactions is crucial.
If you have 1000 grams of water, dividing by the molar mass of water (18.0 g/mol), you get the number of moles. It's a straightforward calculation that's essential in chemistry.
Avogadro's Number
Avogadro's Number is a fundamental constant in chemistry, approximately equal to \(6.022 \times 10^{23}\). This number represents the number of atoms, ions, or molecules in one mole of a substance. It's like a bridge connecting the macroscopic measurements we can handle, to the microscopic world we cannot see directly.
Avogadro's Number helps us understand how vast quantities are involved when dealing with chemical substances. For instance:
  • With water, knowing the number of moles, you can multiply it by Avogadro's Number to get the total number of water molecules.
  • It plays a crucial role in calculations requiring the conversion of moles to molecules, atoms, or ions.
Remember, Avogadro's Number is consistent for any element, making it a universal tool for relating particles and moles.
Molecule Count
Counting molecules might seem daunting, given their incredibly tiny size and vast quantity. However, using mole calculations is very effective. Once you know how many moles of a substance you have, determining the number of molecules involves a simple multiplication with Avogadro's Number.
For example, if you've calculated that you have about 55.56 moles of water in a 1 kg bottle of water (from the previous steps), the number of molecules can be found by:
  • Multiplying the number of moles (55.56 moles) by Avogadro's Number (\(6.022 \times 10^{23}\) molecules/mol).
  • This calculation will give you the total number of water molecules in that amount of water.
This ability to calculate molecule counts from bulk matter is transformative in fields like chemistry and biology, where exact quantities are vital for reactions and applications.

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Most popular questions from this chapter

(a) Show that a projectile with mass \(m\) can "escape" from the surface of a planet if it is launched vertically upward with a kinetic energy greater than \(m g R_{p},\) where \(g\) is the acceleration due to gravity at the planet's surface and \(R_{\mathrm{p}}\) is the planet's radius. Ignore air resistance. (See Problem \(18.70 .\) ) (b) If the planet in question is the earth, at what temperature does the average translational kinetic energy of a nitrogen molecule (molar mass \(28.0 \mathrm{~g} / \mathrm{mol}\) ) equal that required to escape? What about a hydrogen molecule (molar mass \(2.02 \mathrm{~g} / \mathrm{mol}\) )? (c) Repeat part (b) for the moon, for which \(g=1.63 \mathrm{~m} / \mathrm{s}^{2}\) and \(R_{\mathrm{p}}=1740 \mathrm{~km}\) (d) While the earth and the moon have similar average surface temperatures, the moon has essentially no atmosphere. Use your results from parts (b) and (c) to explain why.

(a) Calculate the mass of nitrogen present in a volume of \(3000 \mathrm{~cm}^{3}\) if the gas is at \(22.0^{\circ} \mathrm{C}\) and the absolute pressure of \(2.00 \times 10^{-13}\) atm is a partial vacuum easily obtained in laboratories. (b) What is the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the \(\mathrm{N}_{2}\) ?

You blow up a spherical balloon to a diameter of \(50.0 \mathrm{~cm}\) until the absolute pressure inside is 1.25 atm and the temperature is \(22.0^{\circ} \mathrm{C}\). Assume that all the gas is \(\mathrm{N}_{2},\) of molar mass \(28.0 \mathrm{~g} / \mathrm{mol}\). (a) Find the mass of a single \(\mathrm{N}_{2}\), molecule. (b) How much translational kinetic energy does an average \(\mathrm{N}_{2}\) molecule have? (c) How many \(\mathrm{N}_{2}\) molecules are in this balloon? (d) What is the total translational kinetic energy of all the molecules in the balloon?

A person at rest inhales \(0.50 \mathrm{~L}\) of air with each breath at a pressure of 1.00 atm and a temperature of \(20.0^{\circ} \mathrm{C}\). The inhaled air is \(21.0 \%\) oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of \(2000 \mathrm{~m}\) but the temperature is still \(20.0^{\circ} \mathrm{C}\). Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report "shortness of breath" at high elevations.

The dark area in Fig. \(\mathbf{P} 18.83\) that appears devoid of stars is a dark nebula, a cold gas cloud in interstellar space that contains enough material to block out light from the stars behind it. A typical dark nebula is about 20 light-years in diameter and contains about 50 hydrogen atoms per cubic centimeter (monatomic hydrogen, not \(\mathrm{H}_{2}\) ) at about \(20 \mathrm{~K}\). (A lightyear is the distance light travels in vacuum in one year and is equal to \(\left.9.46 \times 10^{15} \mathrm{~m} .\right)\) (a) Estimate the mean free path for a hydrogen atom in a dark nebula. The radius of a hydrogen atom is \(5.0 \times 10^{-11} \mathrm{~m}\). (b) Estimate the rms speed of a hydrogen atom and the mean free time (the average time between collisions for a given atom). Based on this result, do you think that atomic collisions, such as those leading to \(\mathrm{H}_{2} \mathrm{~mol}-\) ecule formation, are very important in determining the composition of the nebula? (c) Estimate the pressure inside a dark nebula. (d) Compare the rms speed of a hydrogen atom to the escape speed at the surface of the nebula (assumed spherical). If the space around the nebula were a vacuum, would such a cloud be stable or would it tend to evaporate? (e) The stability of dark nebulae is explained by the presence of the interstellar medium (ISM), an even thinner gas that permeates space and in which the dark nebulae are embedded. Show that for dark nebulae to be in equilibrium with the ISM, the numbers of atoms per volume \((N / V)\) and the temperatures \((T)\) of dark nebulae and the ISM must be related by $$ \frac{(N / V)_{\text {nebula }}}{(N / V)_{\text {ISM }}}=\frac{T_{\text {ISM }}}{T_{\text {nebula }}} $$ (f) In the vicinity of the sun, the ISM contains about 1 hydrogen atom per \(200 \mathrm{~cm}^{3} .\) Estimate the temperature of the ISM in the vicinity of the sun. Compare to the temperature of the sun's surface, about \(5800 \mathrm{~K}\). Would a spacecraft coasting through interstellar space burn up? Why or why not?
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