Calculate the freezing point of a solution containing 0.52 g glucose (C6H12O6) dissolved in 80.20g water. For water Kf = 1.86 K Kg mol-1

Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM

Expert Answer:

ΔTf = Kf m

ΔTf = [1.86 x (0.52/180 ) x 1000]   / 80.20       

ΔTf = 0.067 K
Tf (solvent) - (solution)  = 0.067 K
Tf (solution) = 273.15 K-0.067 K
Tf (solution) = 273.083K

Answered by  | 4th Jun, 2014, 03:23: PM