Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K. (kf for water = 1.86 K kg mol⁻¹)

Change in freezing point, ΔTf = kf × m
ΔTf/kf = 2/1.86 = 1.075 mole/kg
m = number of moles of solute/mass of solvent
1.075 = n/1
n = 1.075 moles
Since KCl dissociates into two ions (K⁺ and Cl⁻), the number of particles is twice the number of moles of KCl.
Therefore, moles of KCl = 1.075 / 2 = 0.5375 moles
The molar mass of KCl is approximately 74.55 g/mol.
Mass of KCl = moles × molar mass = 0.5375 moles × 74.55 g/mol = 40.06 g
Therefore, 40.06 g of KCl must be added to 1 kg of water to depress the freezing point by 2K.