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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 | 04 Jun, 2014, 01:23: PM
answered-by-expert 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 | 04 Jun, 2014, 03:23: PM

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