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Chapter 6: Problem 103

When working with a mixture of gases, it is sometimes convenient to use an apparent molar mass (a weightedaverage molar mass). Think in terms of replacing the mixture with a hypothetical single gas. What is the apparent molar mass of air, given that air is \(78.08 \% \mathrm{N}_{2}\) \(20.95 \% \mathrm{O}_{2,0.93 \%} \mathrm{Ar}_{\left(\text {and } 0.036 \% \mathrm{CO}_{2},\text { by volume? } \right.}\)

Short Answer

Expert verified
The apparent molar mass of air is the sum of each component's molar mass times its fraction in the mixture. This is calculated as \((28 g/mol \times 0.7808) + (32 g/mol \times 0.2095) + (40 g/mol \times 0.0093) + (44 g/mol \times 0.00036)\), giving you the apparent molar mass of air in g/mol. Calculate this operation to find the numeric answer.

Step by step solution

01

Identify the molar masses of the individual gases

Refer to the periodic table or other reliable source to find the molar masses of N2 (28 g/mol), O2 (32 g/mol), Ar (40 g/mol), and CO2 (44 g/mol).
02

Convert the percentage composition to a decimal

\(\mathrm{N}_{2}\) makes up 78.08% of air, which is 0.7808 in decimal form. Similarly, \(\mathrm{O}_{2}\) is 20.95% = 0.2095, \(\mathrm{Ar}\) is 0.93% = 0.0093 and \(\mathrm{CO}_{2}\) is 0.036% = 0.00036.
03

Use the percentage composition to find the contribution to the apparent molar mass from each component

For each gas, multiply its molar mass by its percentage (in decimal form) in the mixture. For N2, it will be \(28 \times 0.7808\) g/mol. Repeat the process for each gas.
04

Sum up the contributions to find the total apparent molar mass

After calculating contributions for each component, sum them to get the final apparent molar mass of air.

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Most popular questions from this chapter

Explain why the height of the mercury column in a barometer is independent of the diameter of the barometer tube.

The heat required to sustain animals while they hibernate comes from the biochemical combustion of fatty acids, such as arachidonic acid, \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) What volume of air, measured at \(298 \mathrm{K}\) and \(1.00 \mathrm{atm}\) is required to burn \(2.00 \mathrm{kg} \mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2} ?\) Air is approximately \(78.1 \% \mathrm{N}_{2}\) and \(20.9 \% \mathrm{O}_{2},\) by volume. Other gases make up the remaining \(1.0 \%\)

What are the ratios of the diffusion rates for the pairs of gases (a) \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2} ;\) (b) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{D}_{2} \mathrm{O}\) \(\left(\mathrm{D}=\text { deuterium, i.e., }_{1}^{2} \mathrm{H}\right) ;\) (c) \(^{14} \mathrm{CO}_{2}\) and \(^{12} \mathrm{CO}_{2}\) (d) \(^{235} \mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6} ?\)

What is the volume, in liters, occupied by a mixture of 15.2 \(\mathrm{g} \mathrm{Ne}(\mathrm{g})\) and \(34.8 \mathrm{g} \mathrm{Ar}(\mathrm{g})\) at 7.15 atm pressure and \(26.7^{\circ} \mathrm{C} ?\)

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