In: Chemistry
A sample of glucose (C6H12O6) of mass 8.44 g is dissolved in
2.11 kg water. What is the
freezing point of this solution? The freezing point depression
constant, Kf, for water is
1.86°C/mol.
The depression in freezing point (ΔT) for solutions of non-ionic compounds like glucose is given by:
where:
Kf - freezing point depression constant
m - molality of the solution
Here glucose is the solute and water is the solvent.
The molality of a solution is found as follows:
The molar mass of glucose is 180.16 g/mol
Therefore the number of moles of glucose present in 8.44 g is:
number of moles = 8.44/180.16 = 0.0468 moles
Mass of solvent = 2.11 kg
Therefore molality of the solution is:
Given:
Kf = 1.86 °C/m
The depression in freezing point (ΔT) is:
The normal freezing point of water is 0 °C.
Therefore the depressed freezing point of the solution is (0-0.0413) °C, which equals -0.0413 °C.
Hence the freezing point of the solution is -0.0413 °C.