An aqueous solution of potassium sulfate (k2so4) has a freezing point of –2.24 oc. what is its molality? (kf = 1.86 oc•m–1)

For the answer to the question above,
in this problem we can use the concept of freezing point depression since we are given the freezing point of the solution.

ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
m = 0.4014
So the answer to your question is m = 0.4014

; lc

B"hf and f-"  is the correct answer took the test on edgenuity and got it right