Question

an aqueous solution of sucrose freezes at -36 degrees C. what is its boiling point? (for water kb=0.512 degrees C/m and kf=1.86 degrees C/m

Answer 1

Expression for elevation in boiling point (ΔTb) can be written as follows:

ΔTb = Kb x m x i

Where,

Kb = ebullioscopic constant = 0.512 degrees C/m

m = molality of the solution

i = van’t Hoff factor (for glucose i =1)

Expression for depression in freezing point(ΔTf) can be written as follows:

ΔTf = Kf x m x i

Where,

Kf = cryoscopic constant = 1.86 degrees C/m

m = molality of the solution

i = van’t Hoff factor (for glucose i =1)

Thus, ratio of ΔTb/ ΔTf can be written as follows:

ΔTb/ ΔTf = Kb / Kf

ΔTb = (Kb / Kf ) x ΔTf

ΔTb = (0.512 degrees C/m / 1.86 degrees C/m) x (0degC – (-36degC))

ΔTb = (0.275) x 36degC

ΔTb = 9.9 degC

Therefore, boiling point of water increases by 9.9degC by addition of sucrose.

Thus, boiling point of sucrose solution will be 100 + 9.9= 109.9 degC

Therefore, aqueous solution of sucrose will boil at 109.9 degC.