Question

How much antifreeze (ethylene glycol) must be added to one liter of water in order to lower the freezing point of water by 25 degrees C?

Answer 1

Answer:

0.8342 grams of ethylene glycol must be added to one liter of water.

Explanation:

Volume of water or solvent = 1 L

Density of water = 1 g/L

Mass of the water = Density ×  Volume  = 1 g/L × 1 L = 1 g

1 g = 0.001 kg

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by ethylene glycol is 1 .

i = 1

$\Delta T_f=25^oC$

$\Delta T_f=i\times K_f\times m$

$Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}$

$25^oC=1\times 1.86 ^oC/(mol/kg)\times \frac{Moles}{0.001 kg}$

Moles of solute (ethylene glycol) = 0.013440860 moles

Mass of 0.013440860 moles of ethylene glycol:

$0.013440860 mol\times 62.07 g/mol=0.8342 g$

0.8342 grams of ethylene glycol must be added to one liter of water.