What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

We will use the expression for freezing point depression ∆Tf
     ∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is 
     ∆Tf = 0C - (-3°C) = 3°C 
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and  one Cl- ion:
     KCl → K+ + Cl- 
Therefore, the molality m of the solution can be calculated as 
     3 = 2 * 1.86 * m
     m = 3 / (2 * 1.86)
     m = 0.80 molal

60.0 - there are 3 significant numbers