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How many moles are in a 1.00-kg bottle of water? How many molecules? The molar mass of water is 18.0 g/mol.

Short Answer

Expert verified
1.00 kg of water contains approximately 55.56 moles and about \(3.34 \times 10^{25}\) molecules.

Step by step solution

01

Convert Mass to Grams

First, we need to convert the mass of the water from kilograms to grams. There are 1000 grams in a kilogram. Thus, 1.00 kg of water is equivalent to 1000 grams.
02

Calculate Moles

The formula to find the number of moles is: \[ \text{moles} = \frac{\text{mass of substance (grams)}}{\text{molar mass of substance (g/mol)}} \] Using this formula and the fact that the molar mass of water is 18.0 g/mol, we plug in the numbers: \[ \text{moles of water} = \frac{1000 \text{ g}}{18.0 \text{ g/mol}} \approx 55.56 \text{ moles} \]
03

Calculate Number of Molecules

Using Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules per mole, we calculate the number of molecules: \[ \text{Number of molecules} = 55.56 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \approx 3.34 \times 10^{25} \text{ molecules} \]

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Key Concepts

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

Molar Mass
The concept of molar mass is essential when dealing with mole calculations. Molar mass represents the mass of one mole of a given substance and is expressed in grams per mole (g/mol).
For any compound, the molar mass is calculated by summing the atomic masses of its constituent elements from the periodic table. For instance, the molar mass of water (Hâ‚‚O) is determined by adding the atomic masses of 2 hydrogen atoms and 1 oxygen atom.
- Hydrogen has an atomic mass of approximately 1.0 g/mol.- Oxygen has an atomic mass of approximately 16.0 g/mol.Therefore, to calculate the molar mass of water:\[ \text{Molar mass of H}_2\text{O} = (2 \times 1.0) + 16.0 = 18.0 \text{ g/mol} \]This value is used to convert between the mass of a substance in grams and the amount in moles, which is crucial in chemical calculations.
Understanding molar mass allows you to measure how many moles you have in a given mass of a substance, just as in the original exercise where 1000 grams of water was converted to approximately 55.56 moles.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry that allows scientists to count the number of atoms, molecules, or particles in a sample. The value is approximately \(6.022 \times 10^{23}\) particles per mole. This incredibly large number makes it possible to scale macroscopic quantities in a way that relates to the atomic or molecular level.
In the context of the original problem, once you have the number of moles of a substance, Avogadro's number lets you determine the total number of molecules present.
For example, using Avogadro's number:
  • You determine that 1 mole of water contains \(6.022 \times 10^{23}\) molecules of water.
  • So, 55.56 moles (as calculated from 1 kg of water) will contain: \[ \text{Number of molecules} = 55.56 \times 6.022 \times 10^{23} \approx 3.34 \times 10^{25} \]
Avogadro's number enables the conversion from moles to molecules, offering a clear pathway to understanding the scale of chemical reactions and compositions.
Chemical Calculations
Chemical calculations involve using quantitative relationships between reactants and products in a chemical reaction. These calculations can determine how substances interact based on mass, volume, and number of particles. They are foundational to chemistry and are essential for predicting reaction yields, understanding solution concentrations, and more.
Key processes in chemical calculations include converting between mass and moles, using molar mass, and employing mole ratios from balanced equations.
The original exercise exemplifies a straightforward chemical calculation:
  • First, you convert mass to moles using the molar mass of a substance, as seen when converting 1000 grams of water using its molar mass of 18.0 g/mol, resulting in 55.56 moles of water.
  • Then, incorporating Avogadro’s number lets you convert those moles into molecules, calculating that 1 kg of water contains approximately \(3.34 \times 10^{25}\) molecules.
These calculations not only apply to simple conversions but extend to complex chemical reactions, determining the amounts and outcomes of reactions and much more. Mastery of this concept provides a solid grounding for any further study in the field of chemistry.

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Most popular questions from this chapter

The \(vapor\) \(pressure\) is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The \(relative\) \(humidity\) is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same temperature, expressed as a percentage. The air is saturated when the humidity is 100%. (a) The vapor pressure of water at 20.0\(^\circ\)C is 2.34 \(\times\) 103 Pa. If the air temperature is 20.0\(^\circ\)C and the relative humidity is 60%, what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in 1.00 m\(^3\) of air? (The molar mass of water is 18.0 g/mol. Assume that water vapor can be treated as an ideal gas.)

Helium gas with a volume of 3.20 L, under a pressure of 0.180 atm and at 41.0\(^\circ\)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 g/mol.

The atmosphere of Mars is mostly CO\(_2\) (molar mass 44.0 g/mol) under a pressure of 650 Pa, which we shall assume remains constant. In many places the temperature varies from 0.0\(^\circ\)C in summer to -100\(^\circ\)C in winter. Over the course of a Martian year, what are the ranges of (a) the rms speeds of the CO\(_2\) molecules and (b) the density (in mol/m\(^3\)) of the atmosphere?

In an evacuated enclosure, a vertical cylindrical tank of diameter \(D\) is sealed by a 3.00-kg circular disk that can move up and down without friction. Beneath the disk is a quantity of ideal gas at temperature \(T\) in the cylinder (Fig. P18.50). Initially the disk is at rest at a distance of \(h\) = 4.00 m above the bottom of the tank. When a lead brick of mass 9.00 kg is gently placed on the disk, the disk moves downward. If the temperature of the gas is kept constant and no gas escapes from the tank, what distance above the bottom of the tank is the disk when it again comes to rest?

A balloon of volume 750 m\(^3\) is to be filled with hydrogen at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa). (a) If the hydrogen is stored in cylinders with volumes of 1.90 m\(^3\) at a gauge pressure of 1.20 \(\times\) 10\(^6\) Pa, how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if both the gas in the balloon and the surrounding air are at 15.0\(^\circ\)C? The molar mass of hydrogen (H\(_2\)) is 2.02 g/mol. The density of air at 15.0\(^\circ\)C and atmospheric pressure is 1.23 kg/m\(^3\). See Chapter 12 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 g/mol) instead of hydrogen, again at 15.0\(^\circ\)C?

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