Answer:
The volume of ethylene glycol, we must add is 8.2 L
Explanation:
Let's apply the formula for the colligative property of Depression of freezing point, to solve this
ΔT = Kf . m . i
Where ΔT = Fussion T° in pure solvent - Fussion T° in solution
Kf = Cryoscopic constant (equal to 1.86 °C kg/mol for the freezing point of water)
m = molality (mol of solute in 1kg of solvent)
i = The number of ions dissolved in solution. As it is a non-electrolytic compound, the i values 1 (Van't Hoff factor)
0°C - (-25°C) = 1.86 kg°C/mol . m
25°C / 1.86 m/kg°C = m
13.4 = mol/kg
As water density is 1 g/ml, let's convert firstly 11L in mL
11L .1000 = 11000mL
In conclusion, we have 11000 g of water.
density = mass / volume
So, if we have 13.4 moles in 1 kg of water, how many moles of ethylene glycol do we have, in 11000 g of water.
11000 g = 11kg
1 kg _____ 13.4 moles of ethylene glycol
11 kg _____ (11 . 13.4)/1 = 147.4 moles
Molar mass of ethylene glycol = 62.07 g/m
Moles . molar mass = mass
147.4 m . 62.07g/m = 9149.1 g
Now we can apply density of ethylene glycol to find out the volume
Density ethylene glycol = ethylene glycol mass / ethylene glycol volume
1.11 g/ml = 9149.1 g / ethylene glycol volume
ethylene glycol volume = 9149.1 g/ 1.11 g/ml
ethylene glycol volume = 8242 mL
8242 mL = 8.2 L
