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.

(no3)⁻n has oxidation number x,o has oxidation number -2.(x*1 -2*3) charge of the nitrate ion, and we know that charge of the nitrate ion =-1.so,(x*1 -2*3) =-1x-6=-1x=+5answer a. +5.