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# 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 $0.04 \% \mathrm{CO}_2$. What is the mass, in grams, of a sample of air that contains exactly one mole of the entities?

Solution

Verified

Find the mass in grams of a sample of air.

(1) First, list the molar masses of $\mathrm{N_{2}}$, $\mathrm{O_{2}}$, Ar, and $\mathrm{CO_{2}}$ as shown below:

• Molar mass of $\mathrm{N_{2}}$ = 28.02 g
• Molar mass of $\mathrm{O_{2}}$ = 32.00 g
• Molar mass of $\mathrm{Ar}$ = 39.95 g
• Molar mass of $\mathrm{CO_{2}}$ = 44.01 g

(2) Then, calculate for the mass of air that contains of each mole of entities as shown below:

Mass of air = $(\dfrac{78.08}{100})$ mol $\mathrm{N_{2}}$ x $\dfrac{\text{28.02 g}}{\text{1 mol}\mathrm{N_{2}}}$ + $(\dfrac{20.95}{100})$ mol $\mathrm{O_{2}}$ x $\dfrac{\text{32.00 g}}{\text{1 mol}\mathrm{O_{2}}}$ + $(\dfrac{0.93}{100})$ mol $\mathrm{Ar}$ x $\dfrac{\text{39.95 g}}{\text{1 mol}\mathrm{Ar}}$ + $(\dfrac{0.04}{100})$ mol $\mathrm{CO_{2}}$ x $\dfrac{\text{44.01 g}}{\text{1 mol}\mathrm{CO_{2}}}$

Mass of air = 28.97 g

Thus, the mass of air in this given problem is 28.97.

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