Hydrogen has two naturally occurring isotopes, \(^{1} \mathrm{H}\) and \(^{2} \mathrm{H}\) . Chlorine also has two naturally occurring isotopes, 35 \(\mathrm{Cl}\) and37 Cl. Thus, hydrogen chloride gas consists of four distinct types of molecules: \(^{1} \mathrm{H}^{35} \mathrm{Cl},^{2} \mathrm{H}^{37} \mathrm{Cl},^{2} \mathrm{H}^{35} \mathrm{Cl},\) 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 hydrogen chloride gas molecules is as follows: \(^{2}\mathrm{H}^{37}\mathrm{Cl}\), \(^{1}\mathrm{H}^{37}\mathrm{Cl}\), \(^{2}\mathrm{H}^{35}\mathrm{Cl}\), and \(^{1}\mathrm{H}^{35}\mathrm{Cl}\) (from lowest to highest rate).

Step by step solution

01

Determine the Molar Masses of Each Molecule

To determine the molar mass of each molecule, we need to add the molar mass of hydrogen and the molar mass of chlorine isotope. The molar masses for each isotope should be provided either in the question or in a periodic table, but for this exercise we will assume they are as follows: - \(^{1}\mathrm{H}\): 1 g/mol - \(^{2}\mathrm{H}\): 2 g/mol - \(^{35}\mathrm{Cl}\): 35 g/mol - \(^{37}\mathrm{Cl}\): 37 g/mol Using these values, we can now determine the molar masses for each of the four molecules: 1. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): 1 + 35 = 36 g/mol 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\): 1 + 37 = 38 g/mol 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\): 2 + 35 = 37 g/mol 4. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): 2 + 37 = 39 g/mol
02

Apply Graham's Law of Effusion

Graham's Law states that the rate of effusion is inversely proportional to the square root of the molar mass. Mathematically, this means that: Rate of effusion \(∝ \dfrac{1}{\sqrt{Molar\ Mass}}\) Now, we can compare the rates of effusion for the four molecules based on their molar masses: 1. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{36}}\) = \(\dfrac{1}{6}\) 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{38}}\) 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{37}}\) 4. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{39}}\) It should be noted that the values don't need to be simplified, as we're only interested in comparing the rates. Since larger molar mass results in a smaller rate of effusion and smaller molar mass results in a larger rate, we can now put the molecules in order of increasing rate of effusion: 1. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\) - the lowest rate (highest molar mass) 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\) 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\) 4. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\) - the highest rate (lowest molar mass)

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Most popular questions from this chapter

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