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 solution.
a. 120.1⁰c
b. 0.518 ⁰c
c. 20.1⁰c
d. -20.1⁰c

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 C

hope this helps!

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.