Problem 47
Question
The aqueous solubility at \(20^{\circ} \mathrm{C}\) of \(\mathrm{Ar}\) at \(1.00 \mathrm{atm}\) is equivalent to \(33.7 \mathrm{mL} \mathrm{Ar}(\mathrm{g}),\) measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with air at 1.00 atm and \(20^{\circ} \mathrm{C}\) ? Air contains \(0.934 \%\) Ar by volume. Assume that the volume of water does not change when it becomes saturated with air.
Step-by-Step Solution
Verified Answer
The molarity of Ar in water that is saturated with air at 1.00 atm and \( 20^{\circ} \mathrm{C} \) is 0.000417 M.
1Step 1: Find volume of Ar in air
Air contains 0.934% Ar by volume. This means, in 1000 mL (or 1 L) of air, there will be 0.934% of 1000 mL = 0.00934 * 1000 mL = 9.34 mL of Ar.
2Step 2: Convert volume of Ar to moles
At STP, 1 mole of any gas occupies 22.4 L or 22400 mL. So, 9.34 mL of Ar is equivalent to \( \frac{9.34}{22400} \) moles = 0.000417 moles of Ar.
3Step 3: Calculate molarity
Molarity is defined as moles of solute (Ar) per liter of solvent (water). Considering the water is saturated with air, 1 L of water will contain 0.000417 moles of Ar. Hence, molarity of Ar in air-saturated water at these conditions would be 0.000417 M.
Key Concepts
Aqueous SolubilityVolume to Moles ConversionAir Composition
Aqueous Solubility
Aqueous solubility refers to the maximum amount of a substance that can dissolve in water at a given temperature and pressure. For gases like argon (Ar), solubility is described in terms of the volume of gas dissolved in a certain volume of water, under specific conditions such as standard temperature and pressure (STP).
Understanding aqueous solubility is crucial when determining molarity in solutions. Solubility can change with temperature and pressure, influencing the amount of gas that dissolves in water.
For example, in the case of argon, at 20°C and 1.00 atm, 33.7 mL of gaseous argon can dissolve in 1 liter of water. It means that under these conditions, water can hold a certain fixed amount of argon gas, informing us about its concentration capacity in a solution.
This knowledge is especially useful for calculating how much argon would remain dissolved when water becomes saturated with air, which is a mix of various gases including argon.
Understanding aqueous solubility is crucial when determining molarity in solutions. Solubility can change with temperature and pressure, influencing the amount of gas that dissolves in water.
For example, in the case of argon, at 20°C and 1.00 atm, 33.7 mL of gaseous argon can dissolve in 1 liter of water. It means that under these conditions, water can hold a certain fixed amount of argon gas, informing us about its concentration capacity in a solution.
This knowledge is especially useful for calculating how much argon would remain dissolved when water becomes saturated with air, which is a mix of various gases including argon.
Volume to Moles Conversion
Converting volume to moles is a key step in determining the molarity of a gas dissolved in a liquid. At standard temperature and pressure (STP), 1 mole of any gas occupies a volume of 22.4 liters or 22400 mL. This conversion factor allows us to change the volume of a gas into its moles.
In this particular case, we're dealing with argon gas, for which we have calculated that 9.34 mL is the volume available in 1 liter of air. To convert this volume to moles, we use the ratio:
By understanding this conversion, we effectively translate the measurement of a gas from a volume that is intuitive to a number of moles, allowing for the calculation of molarity. This method can be applied universally to any gas under STP conditions, making it a critical aspect of practicing chemistry.
In this particular case, we're dealing with argon gas, for which we have calculated that 9.34 mL is the volume available in 1 liter of air. To convert this volume to moles, we use the ratio:
- 9.34 mL divided by 22400 mL, which gives approximately 0.000417 moles.
By understanding this conversion, we effectively translate the measurement of a gas from a volume that is intuitive to a number of moles, allowing for the calculation of molarity. This method can be applied universally to any gas under STP conditions, making it a critical aspect of practicing chemistry.
Air Composition
Air composition is an important factor when evaluating how gases like argon interact with water. Air is a mixture of various gases, primarily nitrogen and oxygen, with argon making up about 0.934% by volume. Understanding this composition helps us realize the proportion present in any given volume of air.
In scenarios where air is saturated in water, knowing the exact percentage of each component allows us to calculate the specific amounts present. For instance, in our example, 0.934% of a liter of air corresponds to 9.34 mL of argon. Subtracting or converting these volumes is vital to understand real quantities and manage the solution preparation process.
This compositional knowledge informs us that air saturation in water will result in a certain consistent availability of argon across similar conditions, critical when you need to plan or predict the behavior of gases in aqueous solutions.
In scenarios where air is saturated in water, knowing the exact percentage of each component allows us to calculate the specific amounts present. For instance, in our example, 0.934% of a liter of air corresponds to 9.34 mL of argon. Subtracting or converting these volumes is vital to understand real quantities and manage the solution preparation process.
This compositional knowledge informs us that air saturation in water will result in a certain consistent availability of argon across similar conditions, critical when you need to plan or predict the behavior of gases in aqueous solutions.
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