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Chapter 5: Problem 80

A sample of an unknown gas effuses in 11.1 min. An equal volume of \(\mathrm{H}_{2}\) in the same apparatus under the same conditions effuses in \(2.42 \mathrm{~min} .\) What is the molar mass of the unknown gas?

Short Answer

Expert verified
The molar mass of the unknown gas is approximately 42.0 g/mol.

Step by step solution

01

Understand the Effusion Rates

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. Mathematically, it's expressed as \(\text{Rate of effusion} \propto \frac{1}{\text{sqrt(Molar Mass)}}\).
02

Set Up the Equation Using Graham's Law

Let \[ M_{\text{unknown}} \] be the molar mass of the unknown gas. According to Graham's law: \[\frac{\text{Rate}_{\text{unknown}}}{\text{Rate}_{\text{H}_2}} = \frac{\text{sqrt}(M_{\text{H}_2})}{\text{sqrt}(M_{\text{unknown}})}\].
03

Relate Effusion Times to Rates

The rates of effusion are inversely proportional to the times taken for effusion: \[\text{Rate}_{\text{unknown}} = \frac{1}{t_{\text{unknown}}}\] \ and \[\text{Rate}_{\text{H}_2} = \frac{1}{t_{\text{H}_2}}\]. Substituting these into the equation from Step 2 gives \[\frac{\frac{1}{t_{\text{unknown}}}}{\frac{1}{t_{\text{H}_2}}} = \frac{\text{sqrt}(M_{\text{H}_2})}{\text{sqrt}(M_{\text{unknown}})}\].
04

Simplify the Equation

Simplify the above equation to get: \[\frac{t_{\text{H}_2}}{t_{\text{unknown}}} = \frac{\text{sqrt}(M_{\text{H}_2})}{\text{sqrt}(M_{\text{unknown}})}\]. Plug in the known values: \[ t_{\text{unknown}} = 11.1 \text{ min} \] and \[ t_{\text{H}_2} = 2.42 \text{ min} \].
05

Calculate the Molar Mass of Hydrogen Gas

The molar mass of \(\text{H}_2 \) is \(\text{M}_{\text{H}_2} = 2 \text{ g/mol}\). Substituting the values into the equation: \[\frac{2.42}{11.1} = \frac{\text{sqrt}(2)}{\text{sqrt}(M_{\text{unknown}})}\].
06

Solve for the Unknown Molar Mass

Square both sides to eliminate the square roots: \[ \frac{(2.42)^2}{(11.1)^2} = \frac{2}{M_{\text{unknown}}}\]. Solve for \[ M_{\text{unknown}} \text{ g/mol} \]: \[ M_{\text{unknown}} = 2 \left( \frac{11.1}{2.42} \right)^2 \approx 42.0 \text{ g/mol} \].

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Key Concepts

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

molar mass calculation
To find the molar mass of an unknown gas, we use Graham's Law of Effusion, which relates the rates of effusion of different gases to their molar masses. The equation for Graham's Law is \(\text{Rate of effusion} \propto \frac{1}{\text{sqrt(Molar Mass)}}\).
In our problem, we already know the time it takes for a known gas, hydrogen (\text{H}_2\text{), to effuse, and we use that information to calculate the molar mass of the unknown gas. For hydrogen, the molar mass is \[ \text{M}_{\text{H}_2} = 2 \text { g/mol}\].
We then set up the equation using the given times of effusion and square both sides to solve for the unknown molar mass.
effusion rates
Effusion is the process where gas particles escape through a small hole. According to Graham's Law, the rate at which a gas effuses is dependent on its molar mass.
The key takeaway here is that lighter gases with smaller molar masses will effuse faster than heavier gases. In our exercise, we had to compare the effusion time of hydrogen (\text{H}_2\text{) }with the unknown gas. Using the given values, \[ t_{\text{unknown}} = 11.1 \text{ min} \text { and}\ t_{\text{H}_2} = 2.42 \text{ min}\], we can relate the time taken for effusion to their respective rates by noting that \(\text{Rate} \propto \frac{1}{\text{time}}\).
inverse proportionality
In Graham's Law, the relationship between the rate of effusion and the molar mass of a gas is an inverse proportionality. This means as one value increases, the other decreases. Specifically, the rate of effusion is inversely proportional to the square root of the molar mass.
The formula we use here is \(\frac{\text{Rate}_{\text{unknown}}}{\text{Rate}_{\text{H}_2}} = \frac{\text{sqrt}(\text{M}_{\text{H}_2})}{\text{sqrt}(\text{M}_{\text{unknown}})}\). We substitute the given time values into this equation and then simplify and solve for the unknown molar mass.
This inverse relationship helps us understand that larger molar mass gases will effuse slower compared to smaller molar mass gases.

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Most popular questions from this chapter

An atmospheric chemist studying the pollutant \(\mathrm{SO}_{2}\) places a mixture of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) in a 2.00 -L container at \(800 . \mathrm{K}\) and \(1.90 \mathrm{~atm}\). When the reaction occurs, gaseous \(\mathrm{SO}_{3}\) forms, and the pressure falls to 1.65 atm. How many moles of \(\mathrm{SO}_{3}\) form?

A sample of air contains \(78.08 \%\) nitrogen, \(20.94 \%\) oxygen, \(0.05 \%\) carbon dioxide, and \(0.93 \%\) argon, by volume. How many molecules of each gas are present in \(1.00 \mathrm{~L}\) of the sample at \(25^{\circ} \mathrm{C}\) and 1.00 atm?

The gravitational force exerted by an object is given by \(F=m g\) where \(F\) is the force in newtons, \(m\) is the mass in kilograms, and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) Use the definition of the pascal to calculate the mass (in \(\mathrm{kg}\) ) of the atmosphere above \(1 \mathrm{~m}^{2}\) of ocean. (b) Osmium \((Z=76)\) is a transition metal in Group \(8 \mathrm{~B}(8)\) and has the highest density of any element ( \(22.6 \mathrm{~g} / \mathrm{mL}\) ). If an osmium column is \(1 \mathrm{~m}^{2}\) in area, how high must it be for its pressure to equal atmospheric pressure? [Use the answer from part (a) in your calculation.]

What is the ratio of effusion rates for the lightest gas, \(\mathrm{H}_{2}\), and the heaviest known gas, UF \(_{6} ?\)

Will the volume of a gas increase, decrease, or remain unchanged with each of the following sets of changes? (a) The pressure is decreased from 2 atm to 1 atm, while the temperature is decreased from \(200^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C} .\) (b) The pressure is increased from 1 atm to 3 atm, while the temperature is increased from \(100^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\). (c) The pressure is increased from 3 atm to 6 atm, while the temperature is increased from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). (d) The pressure is increased from 0.2 atm to 0.4 atm, while the temperature is decreased from \(300^{\circ} \mathrm{C}\) to \(150^{\circ} \mathrm{C}\).
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