Question

Two solutions contain the same amount of water as a solvent. One solution contains 0.5 mol glucose (C6H12O6) while the other one contains 0.2 mol of CaCl2 . If the freezing points of both solutions are the same, determine the vanʹt Hoff i factor for CaCl2 solution. A) 2.75 B) 3 C) 0.75 D) 2.50 E) 1.50

Answer 1

The correct answer is D) 2.50

Step-by-step explanation:

The freezing-point depression (ΔTf) of a liquid solvent caused by the addition of a solute can be expressed by the following equation:

ΔTf= Kf x m x i

Where Kf is a cryoscopic constant (it depends on the solvent), m is the molality of the solution (moles of solute per kg of solvent) and i is the van't Hoff factor (it indicates in how many particles is dissociated the solute in the solution).

As the freezing point of C₆H₁₂O₂ and CaCl₂ solutions are the same and both have the same solvent in the same amount, the expressions are equal:

ΔTf (CaCl₂) = ΔTf (C₆H₁₂O₆)

Kf x m x i(CaCl₂)= Kf x m x i(C₆H₁₂O₆)

Kf is the same of both solution, because the solvent is the same (water), so we can eliminate this term and replace m for moles of solute (because we have the same solvent in the same amount). Furthermore, iC₆H₁₂O₆= 1 because C₆H₁₂O₆ is glucose and it is a non ionic solute. Thus,

0.2 mol x i(CaCl) = 0.5 mol x 1

i(CaCl2)= 0.5 mol/0.2 mol

i(CaCl2)= 2.50

Notice that the result is consistent with the number of ions in which CaCl2 dissociates in water (Ca⁺ + 2 Cl⁻= 3). Remember that for most ionic solutes, the actual van't Hoff factor is lower than the predicted by the number of ions in which it dissociates.