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Chapter 10: Problem 31

Suppose you are given two 1-L flasks and told that one contains a gas of molar mass 30 , the other a gas of molar mass 60 , both at the same temperature. The pressure in flask \(\mathrm{A}\) is \(\mathrm{X} \mathrm{atm}\), and the mass of gas in the flask is \(1.2 \mathrm{~g}\). The pressure in flask B is \(0.5 \mathrm{X} \mathrm{atm}\), and the mass of gas in that flask is \(1.2 \mathrm{~g}\). Which flask contains gas of molar mass 30 , and which contains the gas of molar mass 60 ?

Short Answer

Expert verified
Flask A contains the gas with a molar mass of 30 and flask B contains the gas with a molar mass of 60, as there are twice as many moles in flask A compared to flask B.

Step by step solution

01

1. Calculate the moles of gases for flask A and flask B using the Ideal Gas Law

As we know the mass, molar mass, and pressure of both gases, we can calculate the moles of gas using the formula: n = mass / molar mass For flask A: Pressure P = X atm M = 1.2 g For flask B: Pressure P = 0.5*X atm M = 1.2 g As we don’t know the molar mass yet, we denote this as Molar_mass_A for flask A and Molar_mass_B for flask B. Now we can use the Ideal Gas Law to determine the moles. For flask A: \(n_A = 1.2 / \text{Molar_mass_A}\) For flask B: \(n_B = 1.2 / \text{Molar_mass_B}\)
02

2. Compare the moles of the two gases

We will now compare the moles of gases in both flasks, using the pressure ratio to find the relationship between the moles of gas: \(P_A / P_B = n_A / n_B\) Substitute the given values for P_A and P_B: \(X / (0.5X) = n_A / n_B\) or \(2 = n_A / n_B\) Now we know that there are twice as many moles in flask A compared to flask B.
03

3. Identify the molar mass of each gas

Since flask A has twice the moles of gas as flask B, we can conclude that the molar mass of the gas in flask A is smaller than that in flask B. Therefore, flask A contains the gas with a molar mass of 30, and flask B contains the gas with a molar mass of 60.

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate the molar mass of a gas is essential in chemistry, particularly when working with the Ideal Gas Law. Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in one molecule of the substance.

To find the molar mass from a mass of gas and the number of moles, you use the simple formula:
\[ \text{Molar Mass} = \frac{\text{mass of the gas (in grams)}}{\text{number of moles of the gas}} \]
This formula was applied in our exercise, where the mass was given, and the moles were found using the Ideal Gas Law. Once the amount of substance in moles was determined, a comparison between the two different gases allowed for identification of which flask contains which molar mass of gas. Simplicity in this calculation is key, as it lays the foundation for understanding more complex chemical equations and applications.
Gas Law Applications
The Ideal Gas Law, represented by the equation \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for moles, \(R\) for the ideal gas constant, and \(T\) for temperature, is a crucial tool in chemistry. In our example, we see the Ideal Gas Law in action, demonstrating how it can be used to solve real-world problems involving gases.

One key application of the Ideal Gas Law is determining the amount of gas (in moles) present in a container of known volume and pressure, as was needed in our exercise. This calculation is common in lab settings where the reaction yields need to be calculated or when dealing with pressurized systems. By manipulating the Ideal Gas Law, we can predict how a gas will behave under different conditions, such as changes in pressure or temperature, making it a versatile equation in chemistry and engineering.
Mole Concept
The mole concept is a fundamental pillar in the study of chemistry. One mole is defined as the amount of substance containing as many elementary entities (such as atoms or molecules) as there are atoms in exactly 12 grams of carbon-12. This number is Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities per mole.

The mole allows chemists to convert between the mass of a substance and the number of particles it contains. In the given exercise, the mole concept enables us to bridge the gap between the macroscopic world (grams of a substance) and the microscopic world (number of molecules of the gas). Precise understanding and application of the mole concept are necessary for all quantitative aspects of chemistry, as it is used to measure the amount of substance when it comes to reactions, concentrations, and even the Ideal Gas Law.

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Most popular questions from this chapter

A glass vessel fitted with a stopcock has a mass of \(337.428 \mathrm{~g}\) when evacuated. When filled with \(\mathrm{Ar}\), it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of \(\mathrm{Ne}\) and \(\mathrm{Ar}\), under the same conditions of temperature and pressure, it weighs \(339.076 \mathrm{~g} .\) What is the mole percent of Ne in the gas mixture?

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2}\) ? (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{NaClO}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ClO}_{2}(\mathrm{~g})+2 \mathrm{NaCl}(\mathrm{s}) $$ If you allow \(10.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(1.50 \mathrm{~atm}\) at \(21{ }^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

A scuba diver's tank contains \(0.29 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(2.3\) L. (a) Calculate the gas pressure inside the tank at \(9{ }^{\circ} \mathrm{C}\) (b) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and \(0.95 \mathrm{~atm} ?\)

A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(g)\) originally in a 5.00-L vessel at \(21^{\circ} \mathrm{C}\) is transferred to a \(10.0\) - \(\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C}\). A sample of \(2.35 \mathrm{~g} \mathrm{~N}_{2}(g)\) originally in a 2.50-L vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same \(10.0\) - \(\mathrm{L}\) vessel. (a) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(\mathrm{~g})\) in this vessel? (c) What is the total pressure in the vessel?

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and thus has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) are oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane described above and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?
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