Dry air is essentially a mixture of the following entities: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{Ar},\) and \(\mathrm{CO}_{2} .\) The composition of dry air, in mole percent, is \(78.08 \% \mathrm{N}_{2}, 20.95 \% \mathrm{O}_{2}, 0.93 \% \mathrm{Ar}\) and \(0.04 \% \mathrm{CO}_{2}\). (a) What is the mass, in grams, of a sample of air that contains exactly one mole of the entities? (b) Dry air also contains other entities in much smaller amounts. For example, the mole percent of krypton (Kr) is about \(1.14 \times 10^{-4} \% .\) Given that the density of dry air is about \(1.2 \mathrm{g} / \mathrm{L}\) at room temperature, what mass of krypton could be obtained from exactly one cubic meter of dry air?

Step by step solution

01

Determine the Molar Mass of Each Entity

Use the periodic table to find the molar mass of each entity. N_2: \(2 * 14.007 = 28.014 \, g/mol\), O_2: \(2 * 15.999 = 31.998 \, g/mol\), Ar: \(39.948 \, g/mol\), CO_2: \(12.011 + 2 * 15.999 = 43.009 \, g/mol\)

02

Calculate Total Molar Mass of the Air

Multiply each molar mass by its corresponding mole percent (expressed in decimal form) and sum the results: M_air = \(0.7808 * 28.014 \, g/mol + 0.2095 * 31.998 \, g/mol + 0.0093 * 39.948 \, g/mol + 0.0004 * 43.009 \, g/mol = 28.921 \, g/mol\)

03

Mass of One Mole of Air

The mass of one mole of air would be equal to the total molar mass of the air. Therefore, mass of one mole of air, M = 28.921 g.

04

Calculate Molar Mass of Krypton

Using the periodic table, we find that the molar mass of krypton (Kr) is 83.798 g/mol.

05

Calculate the Mass of Krypton in One Cubic Meter of Air

First calculate the number of moles in one cubic meter of air by dividing the mass of air in one cubic meter (which is Density_air * Volume_air = 1.2 g/L * 1000 L = 1200 g) with the molar mass of air (28.921 g/mol). Number of moles in one cubic meter = 1200 g / 28.921 g/mol = 41.49 mol. Then calculate the mass of Kr in the air by multiplying the number of moles of Kr (which is mole percent of Kr * total number of moles = \(1.14*10^{-6}* 41.49\)) with the molar mass of Kr (83.798 g/mol). Mass_Kr = \(1.14*10^{-6} * 41.49 mol * 83.798 g/mol = 0.005 g\).

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