Question

Calculate the amount of CaCl₂ (molar mass = 111 g mol⁻¹) which must be added to 500 g of water to lower its freezing point by 2 K, assuming CaCl₂ is completely dissociated. (K_f for water = 1.86 K kg mol⁻¹)

Answer 1

Given:

Molar mass of CaCl₂ (M_B) = 111 g/mol

Weight of water (wA)       = 500 g

Kf for water                      = 1.86 K kg/mol

ΔT_f                                   = 2 K

Formula,

`DeltaT_f=ixx(K_fxxw_Bxx1000)/(w_AxxM_B)`

CaCl₂ is an electrolyte which dissociates as,

CaCl₂ → Ca²⁺ + 2Cl

Hence, i for CaCl₂ = 3

Solution:

`DeltaT_f=ixx(K_fxxw_Bxx1000)/(w_AxxM_B)`

`2 =3xx(1.86xxw_Bxx1000)/(500xx111)`

`w_B=(2xx500xx111)/(3xx1.86xx1000)`

`w_B=19.9 g`

Amount of CaCl₂ required = 19.9 g