How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?

The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

The kinetic energy of the liquid increases. hope this !