Question

a solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. what is the freezing point depression of the solvent if the freezing point constant -1.86 C/m ?

Answer 1

The freezing point depression of the solution or pure substance that is added with the solvent is calculated through the equation,
    
  ΔTf = Kfm

where ΔT is the freezing point depression, Kf is the constant for water given to be -1.86°C/m and m is the molality of the solution. 

Molality is calculated through the equation,
    
    m = number of moles solute/ kg of solvent

Calculation of molality is shown below.
    
   m = (21.5 g C6H12O6)(1 mol/180 g) / (0.255 kg)
     m = 0.468 molal

The freezing point depression is then,
    
   ΔTf = (-1.86°C/m)(0.468 m) = -0.87°C

Answer: -0.87°C