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

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explanation:

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