3. 1 gal = 3.785 L (density Ethylene glycol = 1.11 g/ ml)
1 gal of ethylene glycol = 3.785 L = 3.785 L * 1000 ml/L * 1.11 g/ml
So, mass of ethylene glycol = 4201.35 g
Ethylene glycol (C2H6O2 ) = 62 g/ mol
moles of Ethylene glycol = 4201.35 g / 62 g/ mol = 67.764 moles.
1 gal of water = 3.785 L = 3.785 L * 1000 ml/L * 1 g/ml
So, mass of water = 3785 g = 3.785 kg
Molality is moles of solvent in per 1000 g (1 kg) of solvent.
Molality = mol of solute / Mass of solvent (in kg) = 67.764 moles / 3.785 kg = 17.903 m (molal)
16.
Freezing point of this aqueous solution :
There is depression in freezing point, freezing point of water = 273.15 K
depression in freezing point = ΔTF = KF · m · i
KF = molal F.P. depression constant = 1.86 °C/m for water
m = molality, i = 1 (vont hoff factor , no ionizaion )
ΔTF = 1.86 °C/m * 17.903 m *1 = 33.3 C
Freezing point of this aqueous solution = 239.82 K
boiling point of this aqueous solution :
There is elevation in boiling point, boiling point of water = 373.15 K
depression in freezing point = ΔTb = Kb · m · i
KF = molal B.P. elevation constant = 0.51 °C/m for water
m = molality, i = 1 (vont hoff factor , no ionizaion )
ΔTF = 0.51 °C/m * 17.903 m *1 = 9.13 C
Boiling point of this aqueous solution = 382.28 K
15. What is the molality of a solution made from 1 gal of ethylene glycol (density...
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hapter 13 kercise 13.121 11 0 1 Review Constants Periodic An ethylene glycol solution is made using 55.5 g of ethylene glycol (C,H,O,) and diluting to a total volume of 800.0 mL Part A Calculate the freezing point of the solution (Assume a density of 1.11 g/mL. for the solution) Express your answer using three significant figures. ΑΣ 03 ? T- Submit Rest Answer Part Exercise 13.121 Revid An ethylene glycol solution is made using 55.5 g of ethylene glycol...