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Chapter 5: Problem 112

The rate of effusion of a particular gas was measured and found to be \(24.0 \mathrm{~mL} / \mathrm{min}\). Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is \(47.8 \mathrm{~mL} / \mathrm{min}\). What is the molar mass of the unknown gas?

Short Answer

Expert verified
The molar mass of the unknown gas is approximately \(4.039 \mathrm{g/mol}\).

Step by step solution

01

Graham's Law of Effusion

The Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this can be represented as: \[\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\] where \(Rate_{1}\) and \(Rate_{2}\) are the rates of effusion of gas 1 and 2, respectively, and \(M_{1}\) and \(M_{2}\) are their molar masses.
02

Given Variables

We are given: Rate of effusion of unknown gas = \(24.0 \mathrm{~mL/min}\) Rate of effusion of methane gas \(\mathrm{(CH_4)}\) = \(47.8 \mathrm{~mL/min}\) We need to find the molar mass of the unknown gas, and we know the molar mass of methane gas \(\mathrm{(CH_4)}\) = \(12.01 + 4(1.008) = 16.042 \mathrm{~g/mol}\).
03

Substitute values in the formula

Using the Graham's Law of Effusion: \[\frac{24.0}{47.8} = \sqrt{\frac{M_{unknown}}{16.042}}\]
04

Calculate the molar mass of the unknown gas

Square both sides of the equation and solve for the molar mass of the unknown gas: \[\left(\frac{24.0}{47.8}\right)^2 = \frac{M_{unknown}}{16.042}\] \[M_{unknown} = 16.042 \times \left(\frac{24.0}{47.8}\right)^2\] \[M_{unknown} = 16.042 \times \left(\frac{576}{2289.64}\right)\] \[M_{unknown} = 16.042 \times 0.251725\] \[M_{unknown} = 4.039 \mathrm{g/mol}\] The molar mass of the unknown gas is approximately \(4.039 \mathrm{g/mol}\).

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Most popular questions from this chapter

An 11.2-L sample of gas is determined to contain \(0.50\) mole of \(\mathrm{N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a 1.0000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(s)\) : $$ \text { Compound } \stackrel{\mathrm{Hot}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2}\). After passage through the KOH solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a \(0.253-\mathrm{g}\) sample of a compound produced \(31.8 \mathrm{~mL} \mathrm{~N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8\) torr.)

Do all the molecules in a 1 -mole sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at \(273 \mathrm{~K} ?\) Do all molecules in a 1 -mole sample of \(\mathrm{N}_{2}(g)\) have the same velocity at \(546 \mathrm{~K} ?\) Explain.

Complete the following table for an ideal gas. $$ \begin{array}{l|lcc|} & P(\mathrm{~atm}) & V(\mathrm{~L}) & n(\mathrm{~mol}) & T \\ \hline \text { a. } & 5.00 & & 2.00 & 155^{\circ} \mathrm{C} \\ \hline \text { b. } & 0.300 & 2.00 & & 155 \mathrm{~K} \\ \hline \text { c. } & 4.47 & 25.0 & 2.01 & \\ \hline \text { d. } & & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\ \hline \end{array} $$
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