Question

15. What is the molality of a solution made from 1 gal of ethylene glycol (density 1.11 g/ml) and I gal of water (density 1.00 g/ml)? 4 pts 16. Determine the freezing point and boiling point for the aqueous solution in number 15 above. Water is the solvent. 4 pts

Answer 1

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 = ΔT_F = K_F · m · i

K_{F =} molal F.P. depression constant = 1.86 °C/m for water

m = molality, i = 1 (vont hoff factor , no ionizaion )

ΔT_F =  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 = ΔT_b = K_b · m · i

K_{F =} molal B.P. elevation constant = 0.51 °C/m for water​​​​​​​

m = molality, i = 1 (vont hoff factor , no ionizaion )

ΔT_F =  0.51 °C/m * 17.903 m *1 = 9.13 C

Boiling point of this aqueous solution = 382.28 K