Help me solve this problem??

Question

What is the expected boiling point of a solution made by dissolving 14.33g of glycerol (C3H8O3) in 51.00mL of water?

The boiling point elevation constant for water is 0.512°C/molal. Report your answer in units of "degC". (show steps)

Answer 1

∆T = imK is the formula to use for this problem.

∆T = change in boiling point = ?

i = van't Hoff factor for glycerol = 1 since it is a non electrolye

m = molality = mols glycerol / kg water = 14.33 g x 1 mol/92.1 g = 0.156 mol / 0.051 kg = 3.06 m

(this assumes a density for water of 1 g / ml. molar mass glycerol = 92.1 g / mol)

K = boiling point constant for water = 0.512ºC/m

Solving for ∆T we have...

∆T = (1)(3.06 m)(0.512ºC/m)

∆T = 1.57ºC ... this is the CHANGE in the boiling point of water, so the boiling point of the solution is ...

Boiling point = 100º + 1.57 = 101.57ºC (not adjusted for sig. figs.)