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

The correct answer is D) 2.50

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.

the product of the pressure and volume is also constant