What mass of ethylene glycol C2H6O2, the main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car's radiator that freezes at -23C? Assume density for water is exactly 1g/mL.

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This is what i have so far:

Freezing point for water is 0.0degree C and Kf = 1.86C/m.

First we find out the freezing point:
=(-23.0)-(0.0_
=-23.0 degrees C

Calculate target Colligative Molarity:
= -23=1.86 * Cm
= 12.37= Cm.

Now we determine the number of grams needed to produce a solution with Cm= 12.37.

We need to find i* molarity:

I don't know how to determine the i: i know that its an integer equal to the number of particles(ions) present in one formula unit of the solute. If i is greater than 1 the molecules dissociates; and if less than 1 then the molecule associates in solution.

and then i use:

molarity = moles of solute (KNO3)/ kg of solvent (water)

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1. CORRECTION : molarity = moles of solute (C2H6O2)/ kg of solvent (water)

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2. i is 1 in antifreeze.
You are not using molarity, you should be using molality, which is moles of antifreeze/kg of solvent

I don't know how potassium nitrate got in there.

12.37=molesantifreeze/kg water
solve for the moles of antifreeze, then convert that to mass

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bobpursley
3. Before I forget, molarity is the number of moles per liter of solution. Molality is the number of moles per kg of solvent. I think your last statement is incorrect. The i for ethylene glycol is 1; i.e., it doesn't ionize.

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4. SO:

First we find out the freezing point:
=(-23.0)-(0.0_
=-23.0 degrees C

Calculate target Colligative Molarity:
= -23=1.86 * Cm
= 12.37= Cm.

i* molality
i= 1

12.37= 1*m
12.37= m

molality= moles of solute(C2H6O2)/kg of solvent water

12.37m= moles C2H6O2/ 10.0kg water
123.7= moles of C2H6O2

123.7moles C2H6O2 * 62.26g C2H6O2
= 7701.56 g of C2H6O2

I think this number is quite high can someone please double check to see if i got the answer right.

Thank YOu

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5. thats rite

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6. thats not right its n

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7. I don't think thats right because it's not 10 kg of water. You need to divide 10 L by 1000 to get 0.01 mL which is then 0.01 g which is then 0.00001 kg water.

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8. Naomi is incorrect, assuming density is 1g/mL, that's equal to 1kg/L. So 10.0L water = 10.0kg water.