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Chapter 18: Problem 27

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

Short Answer

Expert verified
There are approximately 55.56 moles and 3.34 x 10虏鈦� molecules in 1.00 kg of water.

Step by step solution

01

Convert Mass to Grams

First, we need to convert the mass of the water from kilograms to grams because the molar mass is given in grams per mole. Since there are 1000 grams in a kilogram, we have:\[\text{Mass of water} = 1.00 \text{ kg} \times 1000 \text{ g/kg} = 1000 \text{ g} \]
02

Calculate Moles of Water

Next, use the formula to find the number of moles, which is: \(\text{Number of moles} = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}}\). Using the given molar mass of water (18.0 g/mol):\[\text{Number of moles} = \frac{1000 \text{ g}}{18.0 \text{ g/mol}} \approx 55.56 \text{ moles}\]
03

Calculate Number of Molecules

To find the number of molecules, use Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules per mole. Multiply the number of moles of water by Avogadro's number:\[\text{Number of molecules} = 55.56 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \approx 3.34 \times 10^{25} \text{ molecules}\]

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Key Concepts

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

Molar Mass
Molar mass is a key concept in chemistry that helps us understand the amount of matter present in a substance. It is defined as the mass of one mole of a substance, which is usually expressed in grams per mole (g/mol). Think of molar mass as a bridge that connects the mass of a material to the number of moles it contains.
For example, the molar mass of water (H鈧侽) is 18.0 g/mol. This means that 18 grams of water is equivalent to one mole of water molecules. To put this into perspective:
  • Each water molecule consists of 2 hydrogen atoms and 1 oxygen atom.
  • It combines the atomic masses of hydrogen (about 1.0 g/mol) and oxygen (about 16.0 g/mol).
Thus, the calculation gives us 2 * 1 + 16 = 18 g/mol for water. Understanding the molar mass allows us to convert between the physical mass and the chemical amount of a substance.
Avogadro's Number
Avogadro's number is an essential constant in chemistry, named after the Italian scientist Amedeo Avogadro. It provides the link between the macroscopic world of grams and liters, and the microscopic world of atoms and molecules.
Avogadro's number is approximately \(6.022 \times 10^{23}\) entities per mole. This colossal number allows chemists to count particles by weighing them. It tells us how many atoms or molecules are present in a mole of a substance.
  • This number is used to convert moles of a substance into the number of particles.
  • For example, one mole of water contains approximately \(6.022 \times 10^{23}\) molecules of H鈧侽.
Understanding Avogadro's number is crucial for tasks requiring the counting of molecules or atoms, such as chemical reactions or determining the concentration of solutions.
Molecule Calculation
Calculating the number of molecules in a given mass of substance is a common task in chemistry. This involves understanding the relationships between mass, moles, and Avogadro's number.
To calculate the number of molecules:
  • First, determine the number of moles from the given mass using the formula: \(\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}}\).
  • Then, multiply the number of moles by Avogadro's number to find the total number of molecules.
For instance, if you have a 1 kg bottle of water, which weighs 1000 g, and knowing that the molar mass of water is 18.0 g/mol, you first find there are approximately 55.56 moles of water. By using Avogadro's number, multiplying 55.56 moles by \(6.022 \times 10^{23}\), you can calculate that there are about \(3.34 \times 10^{25}\) water molecules in the bottle. This process highlights the power of chemical calculations to connect macroscopic measurements with microscopic particles.

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Most popular questions from this chapter

Helium gas with a volume of 2.60 \(\mathrm{L}\) , under a pressure of 1.30 atm and at a temperanure of \(41.0^{\circ} \mathrm{C},\) is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol.}\)

The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is 1.67 \(\times 10^{-27} \mathrm{kg} . )\) (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2}\) , where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant (see Example 12.5 of Section \(12.3 ) .\) Use the data in Appendix \(\mathrm{F}\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

(a) Compute the specific heat capacity at constant volume of nitrogen \(\left(\mathrm{N}_{2}\right)\) gas, and compare with the specific heat capacity of liquid water. The molar mass of \(\mathrm{N}_{2}\) is 28.0 \(\mathrm{g} / \mathrm{mol}\) . (b) You warm 1.00 \(\mathrm{kg}\) of water at a constant volume of 1.00 \(\mathrm{L}\) from \(20.0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C}\) in a kettle. For the same amount of beat, how many kilograms of \(20.0^{\circ} \mathrm{C}\) air would you be able to warm to \(30.0^{\circ} \mathrm{C} ?\) What volume (in liters) would this air occupy at \(20.0^{\circ} \mathrm{C}\) and a pressure of 1.00 atm? Make the simplifying assumption that air is 100\(\% \mathrm{N}_{2}\) .

A flask contains a mixture of neon (Ne), krypton (Kr), and radon (Rn) gases. Compare (a) the average kinetic energies of the three types of atoms and; (b) the root-mean-square speeds. (Hint: The periodic table in Appendix D shows the molar mass (in g/mol) of each element under the chemical symbol for that element)

A canister of 1.20 mol of nitrogen gas \((28.0 \mathrm{g} / \mathrm{mol})\) at \(25.0^{\circ} \mathrm{C}\) is left on Jupiter's satellite after completion of a future space mission. Europa has no appreciable atmosphere, and the acceleration due to gravity at its surface is 1.30 \(\mathrm{m} / \mathrm{s}^{2}\) . After some time, the canister springs a small leak, allowing molecules to escape through a small hole. What is the maximum height (in \(\mathrm{km}\) ) above Europa's surface that a \(\mathrm{N}_{2}\) molecule having speed equal to the rms speed will reach if it is shot straight up out of the hole in the canister? Ignore the variation in \(g\) with altitude,
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