How much ethylene glycol (C2H6O2, the major component of antifreeze) must be added to 1 L of water to keep it from freezing at - 15 oC? Kf = 1.86 oC/m.

6.97 grams

Explanation:

T = m * Kf

m (molality) = T/Kf

T = - 15°C = - 15+273 = 258 K

Kf = 1.86°C/m = 1.86°C/m * 1K/1°C = 1.86 K/m

Molality = 258/1.86 = 138.71 mol/kg

Mass of ethylene glycol = molality*mass of water in kilograms*molecular weight of ethylene glycol

1 mole of water = 22.4 L

1 L of water = 1/22.4 = 0.045 mol

Mass of water = 0.045 mol * 18g/mol = 0.81 g = 0.81/1000 = 8.1*10^-4 kg

Mass of ethylene glycol = 138.71*8.1*10^-4*62 = 6.97 grams