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 –20ºc? calculate the boiling point of this water-ethylene glycol mixture. (te density of ethylene glycol is 1.11 g/ml.) kf = 1.86 ºc /m

Answer : The volume of ethylene glycol is, 3.92 liters.

The boiling point of a solution is,

Explanation :

First we have to calculate the molality of ethylene glycol.

Formula used :  

where,

= change in freezing point =

i = Van't Hoff factor = 1   (for non-electrolyte)

= freezing point constant =

m = molality of ethylene glycol = ?

Now put all the given values in the above formula, we get:

Now we have to calculate the mass of water.

Now we have to calculate the moles of ethylene glycol.

Moles of ethylene glycol = Molality × Mass of water

Moles of ethylene glycol = 10.8 mol/kg × 6.50kg = 70.2 mole

Now we have to calculate the mass of ethylene glycol.

Mass of ethylene glycol = Moles of ethylene glycol × Molar mass of ethylene glycol

Mass of ethylene glycol = 70.2 mole × 62.07 g/mole =4357.3 g

Now we have to calculate the volume of ethylene glycol.

The volume of ethylene glycol is, 3.92 liters.

Now we have to calculate the boiling point of solution.

Formula used for Elevation in boiling point :

where,

= boiling point of solution = ?

= boiling point of pure water =

= boiling point constant  =

m = molality  = 10.8 mol/kg

i = Van't Hoff factor = 1   (for non-electrolyte)

Now put all the given values in the above formula, we get the boiling point of a solution.

Therefore, the boiling point of a solution is,

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