An aqueous solution boils at 100.50oC. What is the freezing point of the solution? Kb(H2O) = 0.51 oC/m; Kf(H2O) = 1.86 oC/m. Assume a v'ant Hoff factor = 1.
a. -0.50 oC
b. -1.82 oC
c. -5.50 oC
d. 0.50 oC
Chemistry Help????
2019-03-04 10:59 am
回答 (1)
2019-03-04 1:54 pm
Consider the boiling point elevation of the solution:
ΔTb = i × m × Kb
(100.50 - 100.00)°C = 1 × m × (0.51°C/m)
Molality of the solution, m = 0.50 / 0.51 m = 0.98 m
Consider the freezing point depression of the solution:
ΔTf = i × m × Kf
ΔTf = 1 × (0.98 m) × (1.86 °C/m) = 1.82°C
Freezing point of the solution = (0 - 1.82) °C = -1.82 °C
The answer: b. -1.82 °C
ΔTb = i × m × Kb
(100.50 - 100.00)°C = 1 × m × (0.51°C/m)
Molality of the solution, m = 0.50 / 0.51 m = 0.98 m
Consider the freezing point depression of the solution:
ΔTf = i × m × Kf
ΔTf = 1 × (0.98 m) × (1.86 °C/m) = 1.82°C
Freezing point of the solution = (0 - 1.82) °C = -1.82 °C
The answer: b. -1.82 °C
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