Question

depression in freezing point when 45g of ethyelene glycol is mixed with 600g h20

Answer 1

Molar mass of ethylene glycol = 62 g/mol

Therefore, number of moles of ethylene glycol = 45/62 = 0.73 mol  
Mass of water = 600g/1000  = 0.6 kg

Thus, molality of ethylene glycol = 0.73mol/0.60kg = 1.2 m
Therefore freezing point depression,

∆Tf = Kf x m  where, Kf  is the molal freezing point depression constant.

  = 1.86 K kg mol–1 × 1.2 mol kg –1

   = 2.2 K

also as an extra note freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K
that's the answer hope it helps you