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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.
(Kf of water = 1.86 K kg mol-1 ) Assume that ethylene glycol (CH2OH.CH2OH) does not dissociate or associate in aqueous solution.
Solution
Depression in freezing point, `ΔT_f = (1000 xx K_f xx w)/(W xxM )`
Given, W = 4 kg = 4000 g
Kf = 1.86
ΔTf = 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
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