Calculate the freezing point of a 2.00 M solution of CaCl₂ in water.

Data: freezing point depression constant of water = 1.853 K.kg.mol^-1

Question

Calculate the freezing point of a 2.00 M solution of CaCl₂ in water.

Data: freezing point depression constant of water = 1.853 K.kg.mol^-1

| General Chemistry 2

Accepted Answer

Freezing point depression due to CaCl₂:

ΔT_f = - i x K_f x m

ΔT_f = T_f _{solutionCaCl2} – T_{f water}

i = van’t Hoff i-factor = number of ions/molecules produced per formula unit

m = solution molality (in mol.kg^-1)

K_f = proportionality constant (freezing point depression constant)

CaCl₂ = Ca²⁺ + 2 Cl^- ⇒ i = 3

Molality: m (in mol.kg^-1) = n_CaCl2 (in mol) / m_water (in kg)

n_CaCl2 = M_CaCl2x V_solution

and m_water = d_water x V_solution with d_water = volumetric mass density of water = 1.00 kg.L^-1

⇒ m = M_CaCl2 / d_water = 2.00 mol.kg^-1

Freezing point depression:

ΔT_f = - i x K_f x m = - 3 x 1.853 x 2.00 = -11.1 K = - 11.1°C

⇒ T_f _solution – T_{f water} = - 11.1°C

⇒ T_f _solution – 0 = - 11.1°C

⇒ T_f _solution = - 11.1°C

Final - Question 2 | General Chemistry 2