4.What will be the freezing point of aqueous 0.1m KCl. soln, if Kf for 2  86.1OH K.Kg.mol. (a) C186.0   (b) C372.0  (c) C86.1   (d) C

The freezing point depression, ΔTf, is calculated using the formula:

ΔTf = i * Kf * m

where: i is the van 't Hoff factor, which is the number of ions the compound dissociates into in solution. For KCl, this is 2 (K+ and Cl-). Kf is the cryoscopic constant of the solvent (water in this case), which is given as 1.86 °C/m. m is the molality of the solution, which is given as 0.1 mol/kg.

Substituting these values into the formula gives:

ΔTf = 2 * 1.86 * 0.1 = 0.372 °C

Therefore, the freezing point of the solution is decreased by 0.372 °C. Since the normal freezing point of water is 0 °C, the freezing point of the 0.1 m KCl solution is:

0 °C - 0.372 °C = -0.372 °C

So, the answer is (a) -0.372 °C.