Question

Calculate the mass of \[\ce{CaCl2}\] (molar mass = 111 g mol⁻¹) to be dissolved in 500 g of water to lower its freezing point by 2K, assuming that \[\ce{CaCl2}\] undergoes complete dissociation.

(K_f for water = 1.86 K kg mol⁻¹)

Answer 1

ΔT_f = i k_f molality

`DeltaT_f = ("w of CaCl"_2)/("mol. mass of CaCl"_2) xx 1/("wt of water") xx 1000`

`2k = i xx 1.86 xx w/111 xx 1/500 xx 1000`

here, \[\ce{CaCl2 -> Ca^2+ + 2Cl-}\]

i = 3

`2k = 3 xx 1.86 xx w/111 xx 1000/500`

`w = (2 xx 111 xx 500)/(3 xx 1.86 xx 1000)`

w = 19.89 g