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The density of water is \(1.00 \mathrm{~g} / \mathrm{mL}\) at \(4^{\circ} \mathrm{C}\). How many water molecules are present in \(15.78 \mathrm{~mL}\) of water at this temperature?

Short Answer

Expert verified
5.277 × 10^23 water molecules.

Step by step solution

01

Determine the Mass of Water

First, we need to find the mass of the water since we know its volume. Given that the density of water is \(1.00 \text{ g/mL}\), the mass \(m\) of \(15.78 \text{ mL of water}\) can be calculated using the formula: \(m = \text{density} \times \text{volume}\). Thus, \(m = 1.00 \text{ g/mL} \times 15.78 \text{ mL} = 15.78 \text{ g}\).
02

Calculate the Number of Moles of Water

To find the number of moles of water, use the molar mass of water (\(H_2O\)), which is approximately \(18.02 \text{ g/mol}\). The formula to calculate moles is \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Thus, the moles of water is \(\frac{15.78 \text{ g}}{18.02 \text{ g/mol}} = 0.8759 \text{ moles}\).
03

Find the Number of Water Molecules

Use Avogadro's number \(6.022 \times 10^{23}\) molecules/mol to convert moles to molecules. Multiply the number of moles by Avogadro's number: \(0.8759 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} = 5.277 \times 10^{23} \text{ molecules}\).

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Key Concepts

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

Mass of Water
To find the mass of water, you simply need to know the density of water and its volume. Density is a measure of how much mass is contained in a given volume. At 4°C, water has a density of \[1.00 \text{ g/mL}\]. This means that for every milliliter (mL) of water, there is 1 gram (g) of mass. If you have a specific volume of water, say 15.78 mL, you can calculate the mass. Multiplying the density by the volume will give you the mass:
  • Mass \(= \text{Density} \times \text{Volume} = 1.00 \text{ g/mL} \times 15.78 \text{ mL} = 15.78 \text{ g}\).
This straightforward calculation allows us to easily determine the mass of water, which is a crucial step before moving on to other calculations. Always ensure to use the correct units for density and volume to get accurate results.
Moles of Water
After determining the mass of water, the next step is to calculate the number of moles. Moles are a unit for expressing amounts of a chemical substance. To find the moles of water, we use the molar mass of water.The molar mass of water ( \(H_2O\)) is approximately \(18.02 \text{ g/mol}\). This number tells you that one mole of water weighs 18.02 grams. To find the number of moles from mass, use the formula:
  • Moles \(= \frac{\text{Mass}}{\text{Molar Mass}}\).
For our calculated mass of 15.78 grams, we have:
  • Moles of Water = \(\frac{15.78 \text{ g}}{18.02 \text{ g/mol}}\approx 0.8759 \text{ moles}\).
Understanding this conversion helps in relating mass to the amount of substance in fundamental chemical reactions and processes.
Avogadro's Number
Avogadro's number is a fundamental constant that allows chemists to convert between the number of atoms or molecules and moles. It is defined as \(6.022 \times 10^{23}\) particles per mole. This means that one mole of any substance contains exactly \(6.022 \times 10^{23}\) units (such as atoms, molecules, or ions) of that substance.This number is incredibly useful for dealing with substances on a molecular level. After calculating the number of moles of water, you can find out how many water molecules you have using Avogadro's number.For instance, if you have 0.8759 moles of water, multiply by Avogadro's number to get the total number of water molecules:
  • Molecules = \(0.8759 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} = 5.277 \times 10^{23} \text{ molecules}\).
Understanding and using Avogadro's number allows you to bridge the macroscopic world of grams and liters to the microscopic world of atoms and molecules.

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Most popular questions from this chapter

Avogadro's number has sometimes been described as a conversion factor between amu and grams. Use the fluorine atom \((19.00\) amu) as an example to show the relationship between the atomic mass unit and the gram.

A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium oxide \((\mathrm{CaO})\) to form calcium sulfite \(\left(\mathrm{CaSO}_{3}\right) .\) Calculate the daily mass (in kilograms) of \(\mathrm{CaO}\) needed by a power plant that uses \(6.60 \times 10^{6} \mathrm{~kg}\) of coal per day.

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) \(2 \mathrm{NaHCO}_{3} \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O},(\mathrm{b}) \mathrm{NH}_{3}+\mathrm{HCl} \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}\) (c) \(2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

Air is a mixture of many gases. However, in calculating its molar mass we need consider only the three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is made up of 78.08 percent nitrogen, 20.95 percent oxygen, and 0.97 percent argon, what is the molar mass of air?

For many years, the extraction of gold - that is, the separation of gold from other materials- involved the use of potassium cyanide: \(4 \mathrm{Au}+8 \mathrm{KCN}+\mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{KAu}(\mathrm{CN})_{2}+4 \mathrm{KOH}\) What is the minimum amount of KCN in moles needed to extract \(29.0 \mathrm{~g}\) (about an ounce) of gold?

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