Question

You wish to prepare an antifreeze solution for your car’s radiator from distilled water and ethylene glycol (C2H6O2, MW = 62.07 g/mol) that freezes at -15.00℃. What mass, in kg, of ethylene glycol needs to be added to 3.79 liters of distilled water? For water, Kf=1.86℃/m. List your answer with three significant digits.

Answer 1

There is 1.8961Kg of ethylene glycol needed.

Step-by-step explanation:

Step 1: Given data

Molar mass of ethylene glycol = 62.07 g/mole

T = -15.00 °C

Volume = 3.79 L

Kf=1.86°C/m

Step 2: Calculate molality

ΔT = - Kf * m

with ΔT = -15 - 0 = -15 °C ( we use 0 because that's the normal freezing temperature of water).

with K(f) = 1.86°C /m

m = ΔT / Kf

m = -15°C / -1.86°C/m

m =8.06 mole C2H6O2 / 1 kg water

Step 3: Calculate moles needed for 3.79kg

Since 1L=1kg; 3.79L = 3.79kg

⇒ 8.06 moles/1kg * 3.79kg = 30.5474 moles C2H6O2 needed

Step 4: Calculating mass of C2H6O2

mass of C2H6O2 = Moles of C2H6O2 / Molar mass of C2H6O2

mass of C2H6O2 = 30.5474/62.07 g/mole = 1896.08 grams = 1.8961 Kg

There is 1.8961Kg of ethylene glycol needed.