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If you have a sample of water in a closed container, some of the water will evaporate until the pressure of the water vapor, at \(25^{\circ} \mathrm{C},\) is \(23.8 \mathrm{mm}\) Hg. How many molecules of water per cubic centimeter exist in the vapor phase?

Short Answer

Expert verified
Approximately \(7.71 \times 10^{17}\) molecules per cubic centimeter.

Step by step solution

01

Understand the Relationship

The pressure of water vapor implies the presence of a certain number of molecules in a given volume. We are given the pressure at 25°C as 23.8 mm Hg, which needs to be converted into a number density (molecules per cubic centimeter).
02

Convert Pressure to Atmospheres

First, convert 23.8 mm Hg to atmospheres to use it in further calculations.1 atmosphere = 760 mm HgSo, pressure in atmospheres is:\[P = \frac{23.8 \text{ mm Hg}}{760 \text{ mm Hg/atm}}\approx 0.03132 \text{ atm}\]
03

Use Ideal Gas Law

To find the number of molecules per volume, use the Ideal Gas Law, given by:\[ PV = nRT \]where:- \(P\) is the pressure in atmospheres,- \(V\) is the volume in liters,- \(n\) is the number of moles,- \(R\) is the ideal gas constant \(0.0821 \text{ L atm/mol K}\),- \(T\) is the temperature in Kelvin (298 K for 25°C).
04

Calculate Moles Per Cubic Centimeter

Assume a volume of 1 liter (which equals 1000 cubic centimeters). Rearrange the Ideal Gas Law to solve for \(n\) (the number of moles):\[n = \frac{PV}{RT}\]Substitute the known values:\[n = \frac{(0.03132 \text{ atm})(1 \text{ L})}{(0.0821 \text{ L atm/mol K})(298 \text{ K})}\approx 0.00128 \text{ moles}\]Now convert moles to moles per cubic centimeter:\[0.00128 \text{ moles} \div 1000 \text{ cm}^3 = 1.28 \times 10^{-6} \text{ moles/cm}^3\]
05

Convert Moles to Molecules

Convert the moles per cubic centimeter to molecules using Avogadro's number:\[ \text{Number of molecules} = \text{moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \]\[1.28 \times 10^{-6} \text{ moles/cm}^3 \times 6.022 \times 10^{23} \text{ molecules/mole}\approx 7.71 \times 10^{17} \text{ molecules/cm}^3\]
06

Final Step: Conclusion

Therefore, the number of water molecules per cubic centimeter in the vapor phase at 25°C and 23.8 mm Hg is approximately \(7.71 \times 10^{17}\).

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Key Concepts

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

Molecular Number Density
Molecular number density is a measure of how many molecules are present in a specific volume of space. It is particularly useful in understanding gases and their behavior. Imagine you have a container filled with a gas. The molecular number density tells you how crowded those molecules are inside the container. This concept is crucial when we want to predict how a gas will behave under different conditions.

To find the molecular number density, you often start by using the Ideal Gas Law. This law helps relate pressure, volume, and temperature of a gas to the number of moles present. By knowing the number of moles, you can calculate the number of molecules using Avogadro's number. This allows scientists and students alike to determine how packed the molecules are in a given space, whether it's a laboratory experiment or predicting weather patterns.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid form in a closed system. It's important to understand that molecules in a liquid are constantly moving, and some energy-rich molecules escape into the vapor phase above the liquid. This process continues until the rate of evaporation equals the rate of condensation, which results in vapor pressure.

At a given temperature, the vapor pressure is constant for a specific substance. For instance, water at 25°C has a vapor pressure of 23.8 mm Hg. This means that in a closed container, the vapor has enough force to maintain this level of pressure. Understanding vapor pressure allows you to predict how substances will evaporate and condense under various conditions. It's a crucial factor in fields ranging from meteorology to cooking.
Unit Conversion
Converting units is a fundamental skill in science, allowing you to compare and compute various measures easily. Units are like different languages that describe quantities, such as length, time, or temperature. Sometimes, you need to convert one unit to another to perform calculations effectively.

For instance, when working with pressure in scientific problems, you might need to convert from millimeters of mercury (mm Hg) to atmospheres (atm) to use the Ideal Gas Law. The conversion factor here is: 1 atm equals 760 mm Hg. Being proficient in unit conversions ensures accuracy and consistency in your calculations and enhances your understanding of scientific principles.
Avogadro's Number
Avogadro's number is a constant that indicates how many particles, often molecules or atoms, are in one mole of a substance. It's approximately equal to 6.022 x 10^23. This number is astonishingly large because even a small amount of substance contains an immense number of particles.

In practical terms, Avogadro's number allows chemists to convert between the mass of a substance and the number of molecules. If you know the number of moles of a substance, simply multiply by Avogadro's number to find out how many molecules it contains.
  • This concept bridges the macroscopic world, where we measure substances in grams, with the microscopic world of molecules and atoms.
  • It's key to solving problems in chemistry and physics where you need to know the actual number of particles involved.
Understanding Avogadro's number equips you with the power to make sense of matter at a fundamental level.

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Most popular questions from this chapter

In the text, it is stated that the pressure of 4.00 mol of \(\mathrm{Cl}_{2}\) in a \(4.00-\mathrm{L}\) tank at \(100.0^{\circ} \mathrm{C}\) should be 26.0 atm if calculated using the van der Waals equation. Verify this result, and compare it with the pressure predicted by the ideal gas law.

Indicate which represents the higher pressure in each of the following pairs: (a) 534 mm Hg or 0.754 bar (b) 534 mm Hg or 650 kPa (c) 1.34 bar or \(934 \mathrm{kPa}\)

A Chlorine gas \(\left(\mathrm{Cl}_{2}\right)\) is used as a disinfectant in municipal water supplies, although chlorine dioxide \(\left(\mathrm{ClO}_{2}\right)\) and ozone are becoming more widely used. \(\mathrm{ClO}_{2}\) is a better choice than \(\mathrm{Cl}_{2}\) in this application because it leads to fewer chlorinated by-products, which are themselves pollutants. (a) How many valence electrons are in \(\mathrm{ClO}_{2} ?\) (b) The chlorite ion, \(\mathrm{ClO}_{2}^{-}\), is obtained by reducing ClO \(_{2} .\) Draw a possible electron dot structure for \(\mathrm{ClO}_{2}^{-} .\) (Cl is the central atom.) (c) What is the hybridization of the central Cl atom in \(\mathrm{ClO}_{2}^{-} ?\) What is the shape of the ion? (d) Which species has the larger bond angle, \(\mathbf{O}_{3}\) or \(\mathrm{ClO}_{2}^{-} ?\) Explain briefly. (e) Chlorine dioxide, \(\mathrm{ClO}_{2},\) a yellow-green gas, can be made by the reaction of chlorine with sodium chlorite: \(2 \mathrm{NaClO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})+2 \mathrm{ClO}_{2}(\mathrm{g})\) Assume you react \(15.6 \mathrm{g}\) of \(\mathrm{NaClO}_{2}\) with chlorine gas, which has a pressure of \(1050 \mathrm{mm}\) Hg in a \(1.45-\mathrm{L}\). flask at \(22^{\circ} \mathrm{C}\). What mass of \(\mathrm{ClO}_{2}\) can be produced?

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) Place the three gases in order of increasing rms speed at \(298 \mathrm{K}\) (b) If you mix \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio and NO has a partial pressure of \(150 \mathrm{mm} \mathrm{Hg}\) what is the partial pressure of \(\mathrm{O}_{2} ?\) (c) After reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is complete, what is the pressure of \(\mathrm{NO}_{2}\) if the NO originally had a pressure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

A helium-filled balloon of the type used in longdistance flying contains \(420,000 \mathrm{ft}^{3}\left(1.2 \times 10^{7} \mathrm{L}\right)\) of helium. Suppose you fill the balloon with helium on the ground, where the pressure is 737 mm \(\mathrm{Hg}\) and the temperature is \(16.0^{\circ} \mathrm{C}\). When the balloon ascends to a height of 2 miles, where the pressure is only \(600 .\) mm Hg and the temperature is \(-33^{\circ} \mathrm{C},\) what volume is occupied by the helium gas? Assume the pressure inside the balloon matches the external pressure. Comment on the result.

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