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.)

Around 2.0 L of ethylene glycol needs to be added to the car radiator

Explanation:

The depression in freezing point ΔTf of a solution is directly proportional to its molality (m), i.e.

From the given information:

= freezing pt of solution = -10.0 C

= freezing pt of pure solvent = 0 C

Kf = freezing pt depression constant = 1.86 C/m

i = 1 for ethylene glycol antifreeze

Volume of water = 6.50 L = 6500 ml

Density of water = 1.00 g/ml

Therefore mass of water =

Molar mass of ethylene glycol = 62 g/mol

Mass of ethylene glycol needed =

Density of ethylene = 1.11 g/ml

Therefore, volume needed =

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890