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Chapter 5: Problem 41

Some chambers used to grow bacteria that thrive on \(\mathrm{CO}_{2}\) have a gas mixture consisting of \(95.0 \% \mathrm{CO}_{2}\) and \(5.0 \% \mathrm{O}_{2}\) (mole percent). What is the partial pressure of each gas if the total pressure is \(735 \mathrm{~mm} \mathrm{Hg}\) ?

Short Answer

Expert verified
Answer: The partial pressure of CO2 in the gas mixture is 698.25 mm Hg, and the partial pressure of O2 is 36.75 mm Hg.

Step by step solution

01

Calculate the mole fractions of each gas

Mole fraction is given as a percentage in this problem. Convert these percentages into fractions by dividing the percentage by 100. - mole fraction of CO2 = 95.0% / 100 = 0.95 - mole fraction of O2 = 5.0% / 100 = 0.05
02

Calculate partial pressure of each gas

Use the formula for partial pressure: Partial pressure = mole fraction * total pressure Calculate the partial pressure for CO2 and O2 separately: Partial pressure of CO2 = 0.95 * 735 mm Hg = 698.25 mm Hg Partial pressure of O2 = 0.05 * 735 mm Hg = 36.75 mm Hg
03

Present the results

The partial pressure of CO2 in the gas mixture is 698.25 mm Hg, and the partial pressure of O2 is 36.75 mm Hg.

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