ocr: CHAPTER - 10 Solutions to Exercises Gases 10.74 PV.SHT - : ig-"RT: MPV : Change m3 to L, then caiculate grams (or kg). 2.0x 105 m3 X 103 tm dm? X 1dm 1L 1 2.0x108 L H2 2.02 gH2 X 0.08211 K-mol L atm X 18 atm X 300K 2.0x108 L =l 1.6x107 g 1.6 X 10'kgH : L g=1 molH2 10.75 (a) n= RT PV =l 3.00 atm x 0.08211 K-atm L. atm X 300K 1.00L = 0.122 mol C3Hs(g) (b) 0.590 g 1mL C3Hg () X 1.00 X 103 ml X imolC3He 44.094g = 13.4r moi C3H8(I) (c) A - 1.00 L container holds many more moles (molecules) of C2Ha() because in uie liquid phase the molecules are touching. In the gas phase, the molecules are far apar ...
