Question

Two solutions are prepared using the same solute: solution

Answer 1

Cyclohexane will have the greatest freezing point change.  
The equation to model freezing point depression is:
 Î”Tf = Kf · b · i
 where
 Î”Tf = Freezing point depression
 Kf = cryoscopic constant for the solvent
 b = molarity of the solute
 i = van't Hoff factor of the solute (the number of ions produced per solute
molecule) 
 So let's start by calculating the relative molarity of the two solutions.
 Solution a has 0.27 g and solution b has 0.23 g of the solute, so solution a has 0.27/0.23 = 1.173913043 as much solute.
 t-butanol has a density of 0.775 g/mL, so we have 27.4/0.775 = 35.35 ml of solution a.
 cyclohexane has a density of 0.7781 g/ml, so we have 31.87 ml of solution b.
 Since molarity is moles per volume, divide the relative moles by the volumes. So
 solution a: 1.174 / 35.35 = 0.03321075
 solution b: 1 / 31.87 = 0.031377471
 divide both values by 0.031377471 to get relative concentrations:
 solution a: 0.03321075 / 0.031377471 = 1.058
 solution b: 0.031377471 / 0.031377471 = 1 
 So the relative molarities are rather close with solution a only being about
6% higher. 
 We can ignore the van't Hoff factor since we're using the same solute and we
can assume it's the same for both solvents. 
 Now we need the cryoscopic constant for both solvents. Using available tables we find:
 t-butanol: 8.37°C/mol
 cyclohexane: 20.0°C/mol 
 And here we have a very large difference that totally swamps the mere 6% difference in molarity. For approximately the same molarity of solute, cyclohexane will have over twice the freezing point depression.

Answer 2

The freezing point depression is calculated through the equation,

     ΔTf = (Kf)(m)

where ΔTf is freezing point depression, Kf is the molal freezing point depression and m is the molality of the solute. Using the givens above and the equation,


a. Let x be the molar mass of the solute. The Kf of t-butanol is 9.1 °C/m.
      ΔTf = (9.1)(0.27/x)(1/0.0274)
      ΔTf = 89.67/x

b. Let x be the molar mass of the same solute. Kf value of 20.1 °C/m
     ΔTf = (20.1)(0.23/x)(1/0.0248)
     ΔTf = 186.4/x

The answer would be letter B.