A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium.
What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 7.6 atm?
Mon, 2009-10-26 16:39
Convert 2.0g of Oxygen and 98g of Helium into moles. For example if I use the atomic mass of oxygen as 16 (you use the figure required for your course), then 1 mole O2 weighs 32g and 2.0g = 2.0/32 = 0.0625 moles of oxygen. If He = 4, then 1 mole of He weighs 4 g and 98g = 98/4 = 24.5 moles.
So total number of moles of gas in mixture = 0.0625 + 24.5 = 24.5625 moles
Mole fraction of oxygen = 0.0625/24.5625 = 2.5445 x 10-3
Partial pressure of O2 = mole fraction of O2 x total pressure = 2.5445 x 10-3 x 7.6 atm = 0.01934 atmosphere
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