Question

In addition to osmotic pressure, another colligative property that is of great importance in biological systems is freezing point depression. Numerous plant and animal species contain physiological mechanisms that give them “anti-freeze” protection against damage that would be caused by the freezing of their physiological fluids at temperatures below 0ºC. The most common of these anti-freeze mechanisms are based upon glycerol and sorbitol, two organic molecular compounds that occur naturally in the metabolic cycles of many plants and animals. If 15.0g of sorbitol dissolved in 100.mL of water causes the freezing point of the solution to drop to –1.53ºC, what is the molecular weight of sorbitol?

Answer 1

Mass of sorbital = 15.0 g

Volume of water = 100 mL

Freezing point of the solution = -1.53⁰C

We have to calculate molecular weight of the sorbitol.

For that we use colligative property depression in freezing point and its mathematical expression that relates molality and freezing point constant.

We calculate mass of 100 mL of water by using density of H2O .

Mass of 100 mL H2O = 100 mL x 1.0 g /1 mL = 100 g

Mass of water in kg = 100.0 g x 1kg / 1000 g = 0.100 kg

The formula for freezing point

Delta Tf = m x kf

Here Delta Tf is freezing point, Delta Tf = Freezing point of the pure solvent – freezing point of the solution.

m is molality, kf is freezing point constant of water (1.86 ⁰C/m)

Lets plug given values to get m

(0 ⁰C- (-1.53)⁰C = m x 1.86 ⁰ C/m

molality = 0.823 m

We know molality = moles of solute / mass of solute in kg.

And

Number of moles of solute = Mass of g /molar mass.

By rearranging this equations, we can get

Molar mass = Mass of solute in kg / ( mass of solvent in kg x molality)

= 15.0 g / ( 0.8226 m x 0.1 kg )

= 182.35 g/mol

Therefore molar mass of the sorbitol = 182.35 g /mol