Question

12. If 117 g of salt (NaCI) is added to 1500 g of water, what is the freezing point of the resulting solution(10 pts.)? (Note -NaCl yields two particles per mole, soi 2, and molality moles/kq of solvent.)

trying to learn the steps of this equation, details are also helpful. Thank you so much

Answer 1

DTf = i*Kf*m

DTf = Tf-TS

Tf = freezingpoint of solvent = 0 c

Ts = freezingpoint of solution = X c

i= vanthoff factor of solute = 2

Kf of water = 1.86 c/m

m = molality = (W/M)*(1000/Wt of solvent)

             = (117/58.5)*(1000/1500)

             = 1.33 m

(0-(x)) = 2*1.86*1.33

x = freezingpoint of solution = - 4.95 C