Calculate the freezing point of a solution containing 5.7 % kcl by mass (in water).

5.7% KCl  is 94.3 % water.
Therefore, for 1000 g of water the mass of KCl will be (1000× 5.7)/94.3 = 60.445 grams.
1 mole of KCl is equal to 74.55 g, 
therefore, 60.445 g will be 60.445/74.55 = 0.8108 mole of KCl
Hence, 0.8108 moles of KCl should release twice that number of moles 1.6216 moles ions.
Having 1.6216 moles of KCl ions dissolved in 1000g of water, gives us 1.6216 molar if solution.
Using the freezing point depression constant of water.
dT = Kf (molarity)
dT = (1.86 C/ molar) (1.6216 m)
dT = 3.016 C drop in freezing point
Therefore, it should freeze at - 3.016 Celsius

Mass percent of KCl = 5.7%
 So the Mass percent of water = 100 - 5.7 = 94.3%
 KCl mixed in 1000 grams of water = (5.7 / 94.3) x 1000 = 60.44 grams Calculating the moles of KCl = mass of KCL / molecular weight
  = 60.44 / (K (39.1) + Cl(35.5)) = 60.44 / 74.6 = 0.81 moles and KCl release twice the moles, so 2 x 0.81 = 1.62 moles
 So the molarity M = 1.62,
  Kf is the freezing point depression constant of the solvent, here water
 So Kf = 1.86 °C/m
  Freezing point temperature dT = Kf x M = 1.86 x 1.62 = 3.01 C

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