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Hydrogen has two naturally occurring isotopes, \({ }^{1} \mathrm{H}\) and \({ }^{2}\) H. Chlorine also has two naturally occurring isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Thus, hydrogen chloride gas consists of four distinct types of molecules: \({ }^{1} \mathrm{H}^{35} \mathrm{Cl},{ }^{1} \mathrm{H}^{37} \mathrm{Cl},{ }^{2} \mathrm{H}^{35} \mathrm{Cl}\), and \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}\). Place these four molecules in order of increasing rate of effusion.

Short Answer

Expert verified
The order of increasing rate of effusion for the four distinct types of hydrogen chloride gas molecules is \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}, { }^{1} \mathrm{H}^{37} \mathrm{Cl}, { }^{2} \mathrm{H}^{35} \mathrm{Cl},\) and \({ }^{1} \mathrm{H}^{35} \mathrm{Cl}\).

Step by step solution

01

Calculate the Molar Mass of Each Hydrogen Chloride Gas Molecule

First, find the molar mass of each type of hydrogen chloride molecule by adding the atomic masses of the hydrogen and chlorine isotopes involved. Molar mass of \({ }^{1} \mathrm{H}^{35} \mathrm{Cl} = 1 + 35 = 36\) Molar mass of \({ }^{1} \mathrm{H}^{37} \mathrm{Cl} = 1 + 37 = 38\) Molar mass of \({ }^{2} \mathrm{H}^{35} \mathrm{Cl} = 2 + 35 = 37\) Molar mass of \({ }^{2} \mathrm{H}^{37} \mathrm{Cl} = 2 + 37 = 39\)
02

Apply Graham's Law of Effusion

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, lower molar mass means higher effusion rate, and higher molar mass means lower effusion rate.
03

Put the Molecules in Order of Increasing Rate of Effusion

Now, just compare the molar mass of each hydrogen chloride gas molecule and arrange them in order of increasing rate of effusion. Remember that a lower molar mass corresponds to a higher effusion rate. The order of increasing rate of effusion is: \({ }^{2} \mathrm{H}^{37} \mathrm{Cl} (39) \rightarrow { }^{1} \mathrm{H}^{37} \mathrm{Cl} (38) \rightarrow { }^{2} \mathrm{H}^{35} \mathrm{Cl} (37) \rightarrow { }^{1} \mathrm{H}^{35} \mathrm{Cl} (36)\)

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Key Concepts

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

Molar Mass
The molar mass of a substance is the mass of one mole of that substance. It's calculated by adding up the atomic masses of the atoms in the molecule. This calculation is crucial, especially when dealing with molecules with isotopes. Isotopes are variants of elements with the same number of protons but different numbers of neutrons, affecting their atomic mass.

For instance, in hydrogen chloride (HCl) molecules, hydrogen has isotopes
  • the common hydrogen ( 1 H)
  • deuterium ( 2 H)
Chlorine also has isotopes
  • chlorine-35 ( 35 Cl)
  • chlorine-37 ( 37 Cl).
Calculating the molar mass involves adding these isotopic masses. For example:
  • the molar mass of 1 H 35 Cl is 1 + 35 = 36
  • for 1 H 37 Cl, it is 38
Thus, understanding molar mass is crucial not only for the composition but also for predicting the behavior of gases.
Graham's Law of Effusion
Graham's Law of Effusion provides us with an insight into how gases escape through tiny openings. According to this law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In simpler terms, lighter gas molecules effuse faster than heavier ones.

This can be quite useful when determining which type of gas will escape faster in a scenario involving isotopes. For instance, when we know the molar masses of various hydrogen chloride molecules, we can predict their effusion rates. Provided with the formula:\[\text{{Rate of Effusion}} \propto \frac{1}{\sqrt{\text{{Molar Mass}}}}\]we know a molecule with a higher molar mass, like 2H37Cl (39), will effuse slower than a molecule with a lower molar mass, such as 1H35Cl (36).

This application makes Graham's Law indispensable for chemical processes and scientific experiments involving gas effusion.
Hydrogen Chloride Molecules
Hydrogen chloride molecules are a significant subject of study in chemistry. These molecules are comprised of hydrogen and chlorine atoms, which can vary due to their isotopes. Isotopes are important because they lead to various types of molecules, each with a different mass. Considering isotopes:
  • the presence of 1 H and 2 H (deuterium) results in slight mass differences
  • while 35 Cl and 37 Cl augment these differences even further .
This creates distinct hydrogen chloride molecules, such as:
  • 1 H 35 Cl
  • 1 H 37 Cl
  • 2 H 35 Cl
  • 2 H 37 Cl
The diverse molar masses introduced by these isotopes affect their physical properties, such as how quickly they effuse—a critical aspect explored in Graham’s Law of Effusion. Each unique molecule exhibits characteristics that impact both chemical reactions and practical applications like separation processes in labs.

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Most popular questions from this chapter

A deep-sea diver uses a gas cylinder with a volume of \(10.0 \mathrm{~L}\) and a content of \(51.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(32.6 \mathrm{~g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\).

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(4.75\) atm pressure in a 1.00-L container at \(26^{\circ} \mathrm{C}\) is transferred to a 10.0-L container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25\) atm and \(26^{\circ} \mathrm{C}\) in a 5.00-L container is transferred to this same container. What is the total pressure in the new container?

(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr.

A gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{Kr}\) has a density of \(1.104 \mathrm{~g} / \mathrm{L}\) at 435 torr and \(300 \mathrm{~K}\). What is the mole percent \(\mathrm{O}_{2}\) in the mixture?

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is \(65.0 \mathrm{~L}\) and which contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal \(150.0 \mathrm{~atm} ?\) (d) What would be the pressure of the gas, in \(\mathrm{kPa}\), if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

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