If a protein sample has a molar mass of approximately 14,000 u, what is the expected freezing point depression when 1.00g is dissolved in 50.0mL of distilled water?"
-
delta t = m X Kf
for water Kf = 1.86
m = moles of solute per kg of solvent
moles of protein = 1g/ 14000 g / mole = 7.142 x 10^-05 moles
which are in in .050 kg of water if we assume the density of water = 1 g/ ml
moles in 1 Kg = 7.142 x 10^-0 5 moles x 1.00 L / .050 L = 0.001428571
delta t = 0.001428571 x 1.86 = 0.0027 degrees C
the freezing point will be - 0.0027oC
for water Kf = 1.86
m = moles of solute per kg of solvent
moles of protein = 1g/ 14000 g / mole = 7.142 x 10^-05 moles
which are in in .050 kg of water if we assume the density of water = 1 g/ ml
moles in 1 Kg = 7.142 x 10^-0 5 moles x 1.00 L / .050 L = 0.001428571
delta t = 0.001428571 x 1.86 = 0.0027 degrees C
the freezing point will be - 0.0027oC