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

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

Step-by-step 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* K_f* m`

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

`25^oC=1* 1.86 ^oC/(mol/kg)* (Moles)/(0.001 kg)`

Moles of solute (ethylene glycol) = 0.013440860 moles

Mass of 0.013440860 moles of ethylene glycol:

`0.013440860 mol* 62.07 g/mol=0.8342 g`

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