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At \(25^{\circ} \mathrm{C}\) and under an \(\mathrm{O}_{2}(\mathrm{g})\) pressure of \(1 \mathrm{atm},\) the solubility of \(\mathrm{O}_{2}(\mathrm{g})\) in water is \(28.31 \mathrm{mL} / 1.00 \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) At \(25^{\circ} \mathrm{C}\) and under an \(\mathrm{N}_{2}(\mathrm{g})\) pressure of \(1 \mathrm{atm},\) the solubility of \(\mathrm{N}_{2}(\mathrm{g})\) in water is \(14.34 \mathrm{mL} / 1.00 \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) The composition of the atmosphere is \(78.08 \% \mathrm{N}_{2}\) and \(20.95 \% \mathrm{O}_{2},\) by volume. What is the composition of air dissolved in water expressed as volume percents of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
To obtain the precise volume percentages of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), it is necessary to follow through the calculations described in the three steps above.

Step by step solution

01

Calculate the Amount of Each Gas Dissolved in Water

Under given conditions of temperature and pressure, the respective volume of each gas (\(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\)) dissolved in water can be calculated using their solubility ratios mentioned in the problem. For \(\mathrm{O}_{2}\), it'll be \(0.2095 \times 28.31 \, \mathrm{mL}\) and for \(\mathrm{N}_{2}\), it'll be \(0.7808 \times 14.34 \, \mathrm{mL}\).
02

Determine the Total Volume of Dissolved Gases

The total volume of dissolved gases can be determined by adding the volume of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) obtained in step 1.
03

Calculate the Volume Percentage

The respective volume percentage of each gas (\(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\)) can be determined by dividing the volume of each gas by the total volume from step 2, and then multiplying by 100. Perform these calculations to determine the volume percentages of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\).

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Key Concepts

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

Henry's Law and Gas Solubility
Henry's Law is a principle that helps us understand how gases dissolve in liquids like water. It states that the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of that gas above the liquid. This means, if the pressure of a gas increases, more of it will dissolve in the liquid.

This principle is crucial when calculating gas solubility. In the case of oxygen and nitrogen dissolving in water, using Henry's Law allows us to predict how much of each gas will dissolve based on their surrounding pressures. You can imagine it like pressing down on a balloon; the more pressure you apply, the more the gas inside becomes compacted.

Understanding this helps us determine the solubility ratios for our calculations.
Partial Pressure and Its Role
Partial pressure is the pressure exerted by each individual gas in a mixture of gases. In a mixture like the Earth's atmosphere, each component gas, such as oxygen and nitrogen, contributes to the total pressure. Imagine holding a jar filled with gases, where each type of gas pushes against the walls of the jar to create pressure.

When we consider gases dissolving in a liquid, the partial pressure of each gas directly affects the solubility. For instance, in our example, knowing the partial pressure of oxygen and nitrogen helps us use Henry's Law to determine how much each will dissolve in water.
  • Higher partial pressure of a gas means more of it will dissolve.
  • The lower the partial pressure, the less the gas will dissolve.
This concept is key to solving problems involving dissolved gases because it helps set the conditions required for calculating solubility.
Understanding Dissolved Gases
Dissolved gases are gases that have dissolved into a liquid, like sugar dissolving in coffee. When a gas interacts with a liquid, some of its atoms or molecules are absorbed into the liquid, resulting in a solution.

In water, gases like oxygen and nitrogen can dissolve, forming solutions with specific concentrations. For example, when you solve problems involving dissolved gases, you'll need to know how much of each gas is present. This involves using their solubility in water, which tells us the volume of gas that can dissolve in a liter of water under given conditions.

Once the gases are dissolved, they behave much like any other component in a solution, participating in further chemical processes or remaining as part of the solution.
Calculating Volume Percentage of Gases
Volume percentage is a way to express the concentration of a component in a mixture. Even though we might consider gases as part of the air we breathe, they can also dissolve in water. To figure out the concentration of each gas dissolved, we calculate their volume percentage.

Here's how you do it: after determining the dissolved volume of each gas, you add these volumes to find the total dissolved gas volume. To get the percentage for each gas, divide its volume by the total volume and multiply by 100.
  • This provides a clear picture of the makeup of dissolved gases in water.
  • It tells us what fraction of the total volume each gas represents.
Understanding and calculating the volume percentage is crucial when analyzing mixtures and figuring out how they might behave in natural or experimental conditions.

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Most popular questions from this chapter

Explain the important distinctions between each pair of terms: (a) molality and molarity; (b) ideal and nonideal solution; (c) unsaturated and supersaturated solution; (d) fractional crystallization and fractional distillation; (e) osmosis and reverse osmosis.

A solution is prepared by mixing \(1.28 \mathrm{mol} \mathrm{C}_{7} \mathrm{H}_{16}\) \(2.92 \mathrm{mol} \mathrm{C}_{8} \mathrm{H}_{18},\) and \(2.64 \mathrm{mol} \mathrm{C}_{9} \mathrm{H}_{20} .\) What is the (a) mole fraction and (b) mole percent of each component of the solution?

An aqueous solution is \(6.00 \%\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) by mass, with \(d=0.988 \mathrm{g} / \mathrm{mL} .\) What is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution?

Predict the approximate freezing points of \(0.10 \mathrm{m}\) solutions of the following solutes dissolved in water: (a) \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\text { urea }) ;(\mathrm{b}) \mathrm{NH}_{4} \mathrm{NO}_{3} ;(\mathrm{c}) \mathrm{HCl} ;(\mathrm{d}) \mathrm{CaCl}_{2}\) (e) \(\mathrm{MgSO}_{4} ;\) (f) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (ethanol); \((\mathrm{g}) \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (acetic acid).

At \(25^{\circ} \mathrm{C}\) a \(0.50 \mathrm{g}\) sample of polyisobutylene (a polymer used in synthetic rubber) in \(100.0 \mathrm{mL}\) of benzene solution has an osmotic pressure that supports a \(5.1 \mathrm{mm}\) column of solution \((d=0.88 \mathrm{g} / \mathrm{mL}) .\) What is the molar mass of the polyisobutylene? (For \(\mathrm{Hg}\), \(d=13.6 \mathrm{g} / \mathrm{mL} .)\)

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