How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is -10.°C? (The density of ethylene glycol is 1.11 g/mL. Assume the density of water at -10.°C is 1.00 g/mL.)

2.04L

Explanation:

coldest temperature = -10°C

mass of solvent = 6.50kg

freezing point depression = kb*m

Where kb = molar freezing point depression constant, 1.86

10 = 1.86 * molality of ethylene glycol

10 = 1.86 * moles of ethylene glycol/mass of solvent

10 = 1.86 * moles of ethylene glycol/6.5

10*6.5 = 1.86 * moles of ethylene glycol

Moles of ethylene glycol = 65/1.86

Moles of ethylene glycol = 36.11

36.11 mol * 62.1 g/mol = 2242.43g = 2.24kg

1 L = 1000 mL x 1.11 g/mL = 1100 g = 1.1 kg

2.24 kg / 1.10 kg/L = 2.04 L

answer is a

explanation: