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Chapter 10: Problem 85

Hydrogen has two naturally occurring isotopes, \({ }^{1} \mathrm{H}\) and ${ }^{2} \mathrm{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 effusion rates, from fastest to slowest, for these four hydrogen chloride molecules is: \({ }^{1} \mathrm{H}^{35} \mathrm{Cl} < { }^{2} \mathrm{H}^{35} \mathrm{Cl} < { }^{1} \mathrm{H}^{37} \mathrm{Cl} < { }^{2} \mathrm{H}^{37} \mathrm{Cl}\).

Step by step solution

01

Write down Graham's Law of Effusion Formula

Write down the formula for Graham's Law of Effusion: \[ Rate_1/Rate_2 = \sqrt{M_2/M_1} \] Where \(Rate_1\) and \(Rate_2\) are the effusion rates of two different gases and \(M_1\) and \(M_2\) are their respective molar masses.
02

Calculate molar masses for each type of molecule

Next, calculate the molar masses of the four distinct types of hydrogen chloride molecules: 1. \({ }^{1} \mathrm{H}^{35} \mathrm{Cl}\): 1 (hydrogen) + 35 (chlorine) = 36 g/mol 2. \({ }^{1} \mathrm{H}^{37} \mathrm{Cl}\): 1 (hydrogen) + 37 (chlorine) = 38 g/mol 3. \({ }^{2} \mathrm{H}^{35} \mathrm{Cl}\): 2 (hydrogen) + 35 (chlorine) = 37 g/mol 4. \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}\): 2 (hydrogen) + 37 (chlorine) = 39 g/mol
03

Use Graham's Law of Effusion to determine the order of effusion rates

Since the effusion rate is inversely proportional to the square root of the molar mass, we can determine the order by comparing the square root of the molar masses: 1. \(\sqrt{36}\) = 6 2. \(\sqrt{37}\) ≈ 6.08 3. \(\sqrt{38}\) ≈ 6.16 4. \(\sqrt{39}\) ≈ 6.24 The lower the square root of the molar mass, the faster the effusion rate. Thus, the order of effusion rates, from fastest to slowest, is: \({ }^{1} \mathrm{H}^{35} \mathrm{Cl} ({ 6 }) < { }^{2} \mathrm{H}^{35} \mathrm{Cl} ({ 6.08 }) < { }^{1} \mathrm{H}^{37} \mathrm{Cl} ({ 6.16 }) < { }^{2} \mathrm{H}^{37} \mathrm{Cl} ({ 6.24 })\)

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Most popular questions from this chapter

(a) What conditions are represented by the abbreviation STP? (b) What is the molar volume of an ideal gas at STP? (c) Room temperature is often assumed to be \(25^{\circ} \mathrm{C}\). Calculate the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) pressure. \((\mathbf{d})\) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in \(\mathrm{L}\) -bar \(/ \mathrm{mol}-\mathrm{K}\).

A mixture containing $0.50 \mathrm{~mol} \mathrm{H}_{2}(g), 1.00 \mathrm{~mol} \mathrm{O}_{2}(g)\(, and 3.50 \)\mathrm{mol} \mathrm{N}_{2}(g)$ is confined in a 25.0-L vessel at \(25^{\circ} \mathrm{C}\). (a) Calculate the total pressure of the mixture. (b) Calculate the partial pressure of each of the gases in the mixture.

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

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Table 10.3 shows that the van der Waals \(b\) parameter has units of $\mathrm{L} / \mathrm{mol}$. This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for \(\mathrm{Xe},\) calculate the radius of a Xe atom and compare it to the value found in Figure \(7.7,\) that is, \(140 \mathrm{pm}\). Recall that the volume of a sphere is $(4 / 3) \pi r^{3}$.
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