Answer:
333.7g of antifreeze
Explanation:
Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:
ΔT = Kf × m × i
Where:
ΔT is change in temperature (0°C - -20°C = 20°C)
Kf is freezing point depression constant (1.86°C / m)
m is molality of solution (moles solute / 0.5 kg solvent -500g water-)
i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)
Replacing:
20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1
5.376 = moles solute
As molar mass of ethylene glycol is 62.07g/mol:
5.376 moles × (62.07g / 1mol) = 333.7g of antifreeze.
