Question

Part A: A mixture of He, Ar, and Xe has a total pressure of 2.80 atm . The partial pressure of He is 0.250 atm , and the partial pressure of Ar is 0.450 atm . What is the partial pressure of Xe? Express your answer to three significant figures and include the appropriate units. Part B: A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture? Express your answer to three significant figures and include the appropriate units.

Answer 1

solution Payant- A Totod Pressure = 2.809/m. Partial Pressure of Heifhe) = 0.250 am Posetter Pressure of nr Pnv) = 0.450 cem . Protal = PHet Part Pxe .. Rxe = Ptotal – (Pret Par) E = 2.80- (0.250 +0.450) = 2.1 atm or 2.10 atm (3 sig. fig.) Pxe = 2.10 atm Ans. Paxt B Given thatia Temperature of mixture = 0'C or 273K volume (u) = 18.01 Pressurelp) = 1.00 atm. According to ideal gas equation PV = nRT or pV -n (moles of gas) RT Put all above values in the equation; Total mole (m) 1.00 atmx 18.05 of mixture gousse 0.0821 Loatm. molljer y 273K : no 0.803 moi

moles of He = Total mole of mixture -(mole of N₂ + mole of o, ) = 0.803 - (0- 250 +0.250) = 0.303 mole i molar mass of He = 4 g/mole, so mass of He = 0.303 mole x Ag 121g (3 big fig) mole. mass of He = 1.218 Ans.