The difference between the boiling and freezing point

Question

The difference in the boiling point and freezing point of an aqous solutions of glucose 105°C.how much glucose should be dissolved in 1Kg of water? The Kf= 1.86K Kg/mole.Kb=0.51 K/m.molar mass of glucose is 180

Answer 1

Glucose (the solute) is a non electrolyte, so it will not ionize/dissociate in solution. This makes the van't Hoff factor =1 in the boiling point elevation and freezing point depression equations that follow:

∆T = imK

∆T = change in temperature; i = van't Hoff factor; m = molality = moles glucose/kg solvent; K = freezing point or boiling point constant.

Since the ∆T given is not for either freezing or boiling point, but just a difference between the two, we can set that difference = ∆T_b - ∆T_f .  Thus,...

Or, since bp is 100C and freezing is 0C, that would be 100 degrees difference normally. But this is an additional 5 degrees (105 v. 100), so we can take the actual change as 5 degrees.

(1)(m)(0.51) - (1)(m)(-1.86) = 5 

0.51m -(-1.86m) = 5

2.37 m = 5

m = 2.11 moles glucose/kg of solvent

2.11 moles glucose x 180 g/mole = 380 g glucose

 

You can check this answer by finding the change in boiling point and freezing point of this solution as follows:

For boiling point, ∆T = (1)(2.11)(0.51) = 1.1 degrees

For freezing point, ∆T = (1)(2.11)(-1.86) = -3.9 degrees

The difference between the two = 5 degrees