2017-09-06T18:27:34-04:00
16. An aqueous solution has a normal boiling point of 103.5°C. What is the freezing point of this solution? For water and An aqueous solution has a normal boiling point of 103.5°C. What is the freezing point of this solution? For water and -3.6°C -3.5°C -13 °C -0.96°C
SubmitMy An
1
2017-09-08T07:06:15-0400
Solution: Find the molality of the solution: ΔT = imKb ΔT = 103.5 - 100 = 3.5 oC i = 1 m = molality Kb = boiling point constant = 0.51 m = (3.5)/(1)(0.51) = 6.86 m Now, we find the freezing point of this solution ΔT = - imKf ΔT = - (1)(6.86)(1.86) ΔT = - 12.8 oC ≈ - 13 oC
Answer: - 13 oC
Chris O. For water Kb=0.51 degrees C/m Kf=1.86 degress C/m
2 Answers By Expert Tutors
Don I. answered • 09/07/17
Experienced teacher willing to help all students
2 = (1) (0.51) (m)m = 3.92Δt = (1) (1.86) (3.92)Δt = 7.29 CThe freezing point is -7.3C.
Try it with a van 't Hoff factor of 2, you'll get the same result. Since it is the same solution in both cases, the van 't Hoff factor is the same in each equation.
Find the molality of the solution:
∆T = change in boiling point = 3.5ºC
i = van't Hoff Factor = 1 assuming a non-electrolyte as the solute (it isn't stated)
m = molality = moles solute/kg solvent
Kb = boiling point constant = 0.51
m = (3.5)/(1)(0.51) = 6.86 m
Now substitute this value into the freezing point expression and solve for ∆T:
So, the freezing point of the solution will be -12.8ºC