Question

In: Chemistry

A sample of glucose (C6H12O6) of mass 8.44 g is dissolved in 2.11 kg water. What...

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.

Solutions

Expert Solution

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.


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