What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 23.0 L of water to produce an antifreeze solution with a freezing point of -32.0°C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)

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Answer:

Solution

dT = K * b

K (H2O) = 1.86 0C

b(C2H6O2) = dT/ K = 33.2 / 1.86 = 17.85(mol/kg)

b(C2H6O2)=  n (C2H6O2) / m (H2O)

m (H2O) = V * = 12.2 * 1  = 12.2 (kg)

n (C2H6O2) = b (C2H6O2) * m (H2O) = 17.85 * 12.2 = 217.77 (mol)

n (C2H6O2) = m / M;

 m (C2H6O2) = n * M = 217.77 * (12*2 + 1*6 + 16 *2) = 13501.74 (g) = 13.5 (kg)

V (C2H6O2) = m / = 13.5 / 1.11 =  12.16 (L)

Answer V (C2H6O2)  =  12.16 L

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