1 (259)

Prévia do material em texto

Chapter 13 Solutions 259 (b) Given: 21.5 g CuCl₂ in 4.50 X 10² g water, completely dissociated Find: Tf, Tb Other: 1.86 °C/m; Kb = 0.512 °C/m Conceptual Plan: kgH₂O and then m 1 mol CuCl₂ amount solute (moles) m = 134.45 g CuCl₂ mass solvent (kg) and m, i, Kb = = = Kb X i mᵢ=3 Tb = + Solution: 4.5 X 10² = 0.450 kg and 21.5 X 134.45 1 mol CuCl₂ = 0.159911 mol CuCl₂ then m = amount solute (moles) 0.159911 mol CuCl₂ = 0.355358 m then mass solvent (kg) 0.450 kg = X i X m = X 0.355358 m = 1.98 °C then = = 0.000 °C 1.98 = 1.98 and = Kb X i X m = °C X X 0.355358 = 0.546 then = + = 100.000 + 0.546 °C = 100.546 °C Check: The units (°C) are correct. The magnitude of the answers and 100.5 °C) seems reasonable because the molality of the particles is ~1. The shift in boiling point is less than the shift in freezing point because the constant for boiling is larger than the constant for freezing. (c) Given: 5.5% by mass NaNO₃, completely dissociated Find: Tf, Other: = 1.86 °C/m; Kb = Conceptual Plan: Percent by mass then and mass solute 1 mol mass percent = X 100% mass solution 1000 g 85.00 g then m then m, i, and m, i, Kb Tb amount solute (moles) m = mass solvent (kg) = X mᵢ=2 Tf = = + Solution: mass percent = mass solute X 100%, so 5.5% by mass NaNO₃ means 5.5 g NaNO₃ and mass solution 100.0 5.5 g = 94.5 g water. Then 94.5 X 1 kg = 0.0945 kg and 5.5 X 85.00 1 mol = 0.064706 mol NaNO₃ then m = amount solute (moles) = 0.064706 mol = 0.68472 m then mass solvent (kg) 0.0945 = X X m = 1.86°C X X 0.68472 = 2.5 then = = 0.000 °C 2.5 °C = and = Kb X i = °C X 2 X 0.68472 = 0.70 then = + = + 0.70 °C = Check: The units (°C) are correct. The magnitude of the answers and 100.7 °C) seems reasonable because the molality of the particles is ~1. The shift in boiling point is less than the shift in freezing point because the constant for boiling is larger than the constant for freezing. Copyright © 2017 Pearson Education, Inc.