PS1solutions

# PS1solutions - PROBLEMS 1(a(b 12.15 100 = 6.49 187.15 12.15...

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PROBLEMS 1. (a) 12.15 100 6.49 187.15 ×= (b) 2+ 32 Ni(NO ) Ni 12.15 g 0.06650 182.7 g/mol nn == = ; 3 NO 0.1333 n = 2 HO 175g 9.71 18.02 g/mol n 00671 . 0 13 . 0 07 . 0 71 . 9 06650 . 0 X 2 Ni = + + = + 7. 4 MM CuSO (63.55 32.07 64.00)g/mol 159.62 g/mol =+ + = (a) 12.50 159.62 mol 0.164 mol/L 0.478 L M (b) mass = 0.299 mol/L x 0.283 L x 159.62 g/mol = 13.5 g 9. MM = (96.08 + 10.08 + 32.00 + 56.04)g/mol = 194.2 g/mol molality mass % solvent PPM solute X solvent (c) 0.873 85.5 5 1.45 10 × 0.9845 (d) 0.2560 95.263 4 4.737 10 × 0.995408 15. Consider one liter of solution, weighing 1689 g mass 34 HPO = 0.850 × 1689 g = 1436 g; mass 2 HO = 253 g HPO 1436 g 1 mol 14.7 1.00 L 97.99 g M M = 2 14.7 mol 57.9 0.253 kg H O mm 14.7 0.511 14.7 14.0 X +

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CHAPTER 10 17. MM KOH = 56.11 g/mol (a) Consider one liter of solution, weighing 1050 g KOH n = 1.13 mol = 63.4 g; mass water = 987 g 1.13 mol 1.14 0.987 kg water m == 63.4 g mass % 100 6.04% 1050 g = (b) Consider 100 g of solution (30.0 g KOH, 70.0 g 2 HO ) KOH n = 30.0 g/56.11 g/mol = 0.535 mol solution V = 100 g x I mL/1.29 g = 77.5 mL 0.535 mol 6.90 mol/L 0.0775 L M ; m = 0.535 mol/0.0700 kg 2 HO = 7.64 23. (a) 47 1 atm 3.8 10 /atm 5.0 10 /mmHg 760 mm Hg kM M −− × (b) 7_ 4 5.0 10 /mm Hg 293 mm Hg 1.5 10 CM M × (c) 43 He 1.5 10 mol/L 10.00 L 1.5 10 mol n × () 3 (1.5 10 mol)(0.0821 L atm/mol K)(298 K) 293 760 atm 0.095 L 95 mL V ×⋅ = 29. 4 CCl 483 504 0.958 X ; 10 8 CH 0.042 X = Consider one mole total; MM 4 CCl = 153.8 g/mol; MM 10 8 CH = 128.2 g/mol 4 CCl 1 mol 25.00 g 0.163 mol 153.8 g n =×= 0.042 0.163 x x = + , where 10 8 xn =
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## This homework help was uploaded on 04/08/2008 for the course CHEM 128 taught by Professor All during the Spring '08 term at UConn.

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PS1solutions - PROBLEMS 1(a(b 12.15 100 = 6.49 187.15 12.15...

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