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11 Intermolecular Forces Solutions to Exercises (iii) 3.790 10³ kJ required 2219.97 C₃H₈ 44.096 C₃H₈ C₃H₈ = 75 g C₃H₈ (3.8 10³ kJ required has 2 sig figs and so does the result) 11.93 ; T = 273.15 + 26.0 = 299.15 = 299.2 K; P (vapor) = 78.11g/mol 2.0653g 299.15 K 0.08206 mol K atm 760 1atm torr = 98.660 = 98.7 torr 11.94 Plan. Relative humidity and v.p. of H₂O at given T PH₂O ideal-gas law mol H₂O(g) H₂O molecules. Change °F -> °C, volume of room from ft3 1 L. Solve. °C = 5/9 (68 °F 32) = 20 °C; From Appendix B, v.p. of H₂O at 20 °C = 17.54 torr in air = r.h. v.p. of H₂O/100 = 58 17.54 torr/100 = 10.173 = 10 torr 3 = 2.718 10⁴ = 10⁴ L (The result has 1 sig fig, as does the measurement 8 ft.) n = 10.173 torr 760 torr 0.08206 K atm X 293 K = 15.13 = 2 mol H₂O 15.13 mol H₂O 6.022 10²³ molecules = 9.112 = H₂O molecules 1 mol 333