Question

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘C and Kf is equal to 1.86 ∘C/m.

Answer 1

Tf = i*m*Kf

KNO3----> K+ + NO3^-

vant hoff factor is i= 2

Kf                         = 1.86C⁰/m

Tf = i*m*Kf

-14.5 = 2*m*1.86

m = -14.5/3.72 = 3.897m

mass of water = density * volume

                       1*275 = 275gm

molality       = weight of solute*1000/gram molar mass * weight of solvent in gm

      3.897   = W*1000/101*275

      W        = 3.897*27775/1000 = 108.23gm >>>> answer