What is the freezing point of a solution with 30 grams of urea and 235 mL of water

Question

Assume th density of water is 1g over ml

Answer 1

To calculate freezing point of a solution, we use the equation 

deltaT_f = i x m x K_f

where i is the vast Hoff factor associated with the amount of dissociation of the solute in solution, m is the molality of the solution, and K_f is the molal freezing point depression constant. 

 

Since urea is a covalently bonded compound, it does not dissociate in solution so its i value or want Hoff factor is 1. 

 

To find the molality, with divide moles of solute by kg of solvent. Here, we have 30 grams of solute, which is 0.5 moles, and 235mL of solvent, which is 235 g (since the density of water is 1 g/mL) or 0.235 kg. Therefore the molality = 0.5/0.235 = 2.13 m

 

The Kf value should be given to you or can be easily found in your textbook or some other table/resource from your teacher. Then we substitute all of those values into the equation and multiply to find the change in freezing point! From there, since freezing point of a solution is always depressed, the last step is to subtract that value from the original freezing point of pure water (which can also be looked up pretty easily).

 

Answer 2

Colligative Properties

∆T = imK

∆T = change in freezing point

i = van't Hoff factor = 1 for non electrolyte urea

m = molality = moles solute/kg solvent = 30 g 1 1 mol/60 g /0.235 kg = 2.13 m

K = freezing point constant for water = 1.86

∆T = (1)(2.13)(1.86) = 3.96 degrees

Freezing point of solution will be -3.96 degrees C (-4ºC)