In: Chemistry
An antifreeze protein, with a molecular weight of 17000 Da, was found to be present at a concentration of 10g dm-3 in the serum of an Antarctic fish. calculate the expected depression of the freezing point and compare your result with 1) the observed depression of 0.6K and 2) the depression of 0.0014K observed for 10g dm-3 solution of lysozyme, molecular weight of 14500 Da. for pure water, Lf=6.02kj ml-1 and Tm=273.15 K
Solution:
We know ∆Tf = -Kf x m
Where, ∆Tf = depression of freezing point, Kf = is the freezing point depression constant, and m = molality.
Concentration of the protein = 10g dm-3
=10/17000 mol dm-3 = 5.88x10-4 mol/L
= 5.88x10-4 mol/kg (assuming density of water = 1kg/L)
Therefore, ∆Tf = -Kf x m = -274.86x5.88x10-4 = -0.16 K (since, kf = 274.86 K/m)
Similarly, Concentration of lysozyme = 10g dm-3
=10/14500 mol dm-3 = 6.89x10-4 mol/L
= 6.89x10-4 mol/kg (assuming density of water = 1kg/L)
Therefore, ∆Tf = -Kf x m = -274.86x6.89x10-4 = -0.19 K (since, kf = 274.86 K/m)