A solution of antifreeze contains 145 g of ethylene glycol (C2H6O2) per 550. gram of water. What is the freezing point of this solution? (Kf = 1.86 oC/m)
The freezing point of solution can be calculated as;
ΔTF = KF · b · i
again ,ΔTF = i X KF X (W2 X 1000) / (M2 X W1) ...................(1)
Where
i = van 't Hoff factor = 1
b =molality = (W2/M2) / (W1/1000)
W2 = weight of ethylene glycol =145 g
W1 = Weight of water = 550 g
M2 = Moler mass of ethylene glycol = 62.07 g/mol
KF = cryoscopic constant = 1.86 deg C/m = = 1.86 K·kg/mol
Putting the value in equation (1) we get
ΔTF = (1) X (1.86 X 145 X 1000 ) / (62.07 X 550) = 269700 / 34138.5 = 7.9 K
Again,
ΔTF = TF (pure solvent) − TF (solution) ...............................(2)
Here TF (pure solvent) 273.15
TF (solution)= TF (pure solvent) − ΔTF = 273.15 - 7.9 = 265.25 K = - 7.9 deg C
Freezing point of solution is - 7.9 deg C.
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