Question

Part A What mass of carbon dioxide is present in 1.00 m of dry air at a temperature of 25 °C and a pressure of 705 torr? Express your answer with the appropriate units. View Available Hint(s) Air is a mixture of several gases. The 10 most abundant of these gases are listed here along with their mole fractions and molar masses. Molar mass Component Mole fraction (g/mol) Nitrogen 0.78084 28.013 Oxygen 0.20948 31.998 Argon 0.00934 39.948 Carbon dioxide 0.000375 44.0099 Neon 0.00001818 20.183 Hellum 0.00000524 4.003 Methane 0.000002 16.043 Krypton 0.00000114 83.80 Hydrogen 0.0000005 2.0159 Nitrous oxide 0.0000005 44.0128 * O2, ? kg 1.7 Submit Previous Answers * Incorrect; Try Again; 4 attempts remaining Part B Calculate the mass percentage of oxygen in dry air. Express your answer with the appropriate units. View Available Hint(s) , ? O Units Value Submit

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Answer 1

Qo Part A Given that Nolume of dry air (u) = 1m² Floool Gilms cloool) perature (7225° = 25+273K E 298K 1 760 pressure (P) = 70store=tos atm 20.928 atm (late=760 tory we know that from ideal gast haw PV=nRT n= P = 0.928 at - where - 0.0821 Lata x298K n nor of coles Py pressure, vy Volume 2 no 928 mol 1 T Temperature. 1 24.46 ence, n = 37.94 mol Moles of dry air(n)=37.94 mal we know that molar mass of con=448lmal Nowo mass of co2 = 37.94 exolxuu8lmot =1669.369 Answer = 1669.367X0.CORE 167kg Part-B)→ we know that mass percent of ceae-cass of gas gas = X100 total mass of gasses mass y of oxygen gas a mass of oxyg totoul mass of dry air we know mass of that.. gass = molar fraction of gass x molar mass of gas oxygen - Molar çration of oxu mass of Malay MASS of oxygen -0.20948*31.938)9 = 16.709

In this way we can dry air or mixture mass of calculate zozal of gasses. BY calculation we will find out that total mass of dry air or mixture of gasses = 298 approx) رده كى mass percent of oxygen - mass of oxygen. ?x200 Total mass of dry air 6.708 -X100 0.2310 X100 =23-104] 29 g Hence, onass percentage of oxygen in dry air is [23.10%.