An antifreeze solution is prepared containing 40.0 g of ethylene glycol (molar mass = 62.0 g/mol) in 60.0 g of water. calculate the freezing point of this solutions a.-20.1°c b.0.518°c c.20.1°c d.120.1°c

Kf = 1.86

Number of moles :

Molar mass  ethylene glycol = 62.0 g/mol

40.0 / 62.0 => 0.645 moles 

Molality :

mass of solvent = 60.0 / 1000 => 0.06 kg

m = moles / kg solvent

m = 0.645 / 0.06

= 10.75 m

Δt = Kf x molality

Δt = (- 1.86) x 10.75 => - 20.1ºC

Answer A

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