What is the freezing point of a 0.743m aqueous solution if kci? (report amount to three decimal points)

This is a question about the colligative property known as freezing point depression. Freezing point depression (the amount the normal freezing point of the solvent is decreased) can be calculated with this equation:

ΔT = i Kf m

Where i (the van't Hoff factor) is the degree of dissociation of the solute, Kf is the freezing point depression constant, and m is the molality of the solution.

Here i = 2 (KCl dissociates into 2 ions, K+ and Cl-), Kf = 1.86 C/m (for water), and m = 0.743m).

ΔT = 2 x 1.86 C/m x 0.743m = 2.764C

That means the freezing point of the solution is 2.764C less than the pure solvent (water), making it 0C - 2.764C = -2.764C.

Change in freezing point = -1.86 C/m X 0.743 m X 2 = -2.76 C.. I hope that helps

The answer is c fish