Question

What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 11.0 L of water to produce an antifreeze solution with a freezing point of -25.0°C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)

Answer 1

The volume of ethylene glycol, we must add is 8.2 L

Step-by-step 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