How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.36 ∘c? express the mass in grams to three significant digits?

For this, we make use of the colligative properties of solutions specifically the freezing point depression. It is expressed as follows:

ΔT = Km    where K is a constant and is equal to 1.86 for water and m is the molality in mol/kg

5.36 = 1.86 (m)
m = 2.88 mol /kg

mass of ethylene glycol = 2.88 (1.25) ( 1/ 62.08 ) = 0.0580 g

Hope this helps.