Question

Ethylene glycol is used as an antifreeze agent. Calculate the amount of ethylene glycol to be added to four kilogram of water to prevent it from freezing at −6°C.

(K_f of water = 1.86 K kg mol^-1 ) Assume that ethylene glycol (CH₂OH.CH₂OH) does not dissociate or associate in aqueous solution.

Answer 1

Depression in freezing point, `ΔT_f = (1000 xx K_f xx w)/(W xxM )`

Given, W = 4 kg = 4000 g 

K_f = 1.86

ΔT_f = 0 -(-6°) = 6° C

Mol. mas (M) of ethylene glycol \[\ce{C2H6O2}\] = 62 

∴ Amount of ethylene glycol required, 

`w = (M xx W xx DeltaT_f)/(1000 xx K_f)`

`w = (62 xx 4000 xx 6)/(1000 xx 1.86)`

w = 800 g