A solution is made of 2.05g of sucrose [C12H22O11] dissolved in 9.87g of water . Calculate freezing and boiling points of sucrose solution

Question

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Answer 1

Freezing point depression and boiling point elevation are both colligative properties, meaning they depend on the number of particles present in solution, and not so much on the nature of those particles. The formula we use to calculate the change in freezing point and/or boiling point is...

∆T = imK where ..

∆T = change in temperature

i = van't Hoff factor

m = molality = moles of solute / kg of solvent

K = boiling or freezing constant for the particular solvent

In the current problem, we have the following:

molar mass sucrose = 342 g / mol

∆T = ?

i = 1 for sucrose, as it is a non electrolyte so produces only 1 particle

m = 2.05 g x 1 mol / 342 g = 0.00599 mols / 0.00987 kg water = 0.607 molal (m)

K_freeze = 1.86º/m and K_boil = 0.512º/m

For the freezing point, we have...

∆T = (1)(0.607 m)(1.86º/m) = 1.13º (note: this is the CHANGE in freezing point).

Freezing point of solution = 0º - 1.13º = -1.13º

For the boiling point, we have...

∆T = (1)(0.607 m)(0.512º/m) = 0.311º (note: this is the CHANGE in boiling point)

Freezing point of solution = 100º + 0.311º = 100.311º (not corrected for sig.figs.)