Question

1. A 0.250 g sample of a non‐volatile solid dissolves in 15.0 g of tert‐butanol (Freezing point 25.5 degrees C, Kf=9.1C/m).  The freezing point of the solution is 20.7 degrees C.

a. What is the molality of the solute in the solution?

b. Calculate the molar mass of the solute.

c. The same mass of solute, 0.250 g, is dissolved in 15.0 g of ethylene glycol (Freezing point= -12.7C, Kf=3.11C/m) instead of tert‐butanol.  What is the expected freezing point change of this solution?

Answer 1

dTf = i*Kf*m

DTf = T0-Ts = 25.5-20.7 = 4.8 c
  
i = vanthoff factor of solute = 1

Kf of tert‐butanol = 9.1 c/m

   m = molality = ? m

4.8 = 1*9.1*m

m = 0.527 m

a) molality = 0.527 m

     0.527 = (0.25/x)*(1000/15)

b) molarmass of the compound = x = 31.62g/mol

c.

dTf = i*Kf*m

DTf = T0-Ts = -12.7-x c
  
i = vanthoff factor of solute = 1

Kf of ethylene glycol = 3.11 c/m

   m = molality = (0.25/31.62)*(1000/15) = 0.527 m

x = 1*3.11*0.527

x = 1.64

DTf = -12.7-x = 1.64

x = Freezing point of solution = -14.34 c

change = DTf = -12.7-x = 1.64