Question

QUESTION 1 0.5 points Save Answer A mixture of He, Ne and Ar has a pressure of 7.85 atm. If the Ne has a mole fraction of 0.47 and Ar has a mole fraction of 0.23, what is the pressure of He? 5.5 atm 4.2 atm O 3.7 atm 2.4 atm 1.8 atm QUESTION 2 0.5 points Save Answer A vessel contains a mixture of 30.3 grams of H2 (g) and 13.6 grams of CH4. If the total pressure inside the vessel is measured at 3.30 atm, the partial pressure of H2 (8) must be 49.6 O 0.176 6.30 3.12 52.4 QUESTION 3 0.5 points Saye Answer

Answer 1

The complete step wise calculation is given as below:-

0 It is given that, Total pressure (PT) = 7.85 atm. mole fraction of Ne=Xne = 0.47 mole fraction of Ar = Xar= 0.23 mole fraction of He= Xhe=1- (x Ne tXA) on the = 1 - (0:47 +0.23) = 0.3 or the =0,3 So, the pressure af He will be

Putting all the values in the above equation , the = (013) (7.85) atm = 2.355 atm. or Pressure afte = kre= 2.4 atm. So, the correct option will e @ 2.4 atm. It is given that, Total pressure (Pt) = 3:30 atm mass of H₂ = 30.3 g molecular weight aft2 = 291mol Number of moles afte=30 but more the gimol 30. 39 - 15.15 ma

mass ab Chu = 13.6g. Molecular weight of CH4 = 16g1mal. Number of moles of CH4 = 13.6 g = 0.85 moles Toglmal Total number afmoles (t) = Nua thchu or nr =(15:15 +0.85) moles = 16 moles. nne z = 0.95 mole fraction of H2 = XH2 = 0 16

Partial pressure af He is given by! - tua=XH₂ Pt - putting all the values in the above equation @ PH2=XH2 Py = 0.95 X 3.30 atm = 3.135 atm Partial pressure H2 = kno=3.135 atm. So, the correct option will be ② 3.12 atm.