1.0345 g of palmitic acid dissolved in 8.7545 g of stearic acid. Calculate the freezing point of the solution. The molar mass of palmitic acid is 256.48 g/ mol. The molal freezing point constant , Kf for stearic acid is 4.5 degrees C/ m and pure stearic acid freezes at 69.3 degrees Celsius.
Moles of palmitic acid = mass / molar mass
= 1.0345 / 256.48
= 4.033*10^-3
Molality = moles / mass of solvent (in kg)
= 4.033*10^-3 / ( 8.7545*10^-3)
= 0.461
∆T = kf*m
= 4.5 * 0.461
= 2.0745
T = T° - ∆T
= 69.3 - 2.0745
= 67.23 °C
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