Chem=continued

I'm currently working on a lab in which I have to make the solution boil at 245 degrees F. The solution will be a mix of ethylene glycol and I have to caluculate the volumes of water and solute needed to make the solution.

here's my work so far.
X= mL C2H4(OH)2
175mL-X=mL H20

water boils @ 100 degrees C
my bp is 118.33 degrees C

given:
kb= .515 degrees C kg/mol
d of C2H4(OH)2 = 1.1088 g/mL

delta Tb = (kb*g*i)/GMW*kg

18.33 degrees c = (.515)*(1.1088)*(1)/(62.08g/mol)*(175-X(.9970)/1000)
18.33 = .571032X/(62.08)(175-X)(.9970)/1000
18.33 = 571.032 X/(62.08)(174.475-.9970X)
18.33 = 571.032X/ (1031.408-61.89376X)

now i'm stuck. i don't understand how to get x by itself from here. is what i've done so far correct?
thank you.

I got cross eyed looking at all the numbers. Let me break it down. Also, you don't say how much you want to make. I have assumed 100 g water as the solvent.
Delta T = Kb*m (i is 1).
118.33-100 = 0.515*m
solve for molality.

molality = mols solute/kg solvent = mols solute /0.1
solve for moles solute.

mols solute = grams solute/molar mass solute

If you want to use volume for the ethylene glycol instead of weighing it, then volume = mass desired/density

I hope this helps.

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  1. thank you. i'm confused thought....i tried to use a formula that already had molality in it because i don't have a mass to use, hence the g/GMW kg portion. i need to use 175 mL and make it boil at 245 degrees F.

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  2. Can you make up any quantity, then use 175 mL of that solution or must you make up exactly 175 mL. If the former, the procedure I outlined in my previous post will work ok but you will end up using more chemicals than is necessary AND you will have much more solution that is unused. If the latter, I don't know how to do it except by trial and error.
    Using 118.33-100 = 0.515 m we can determione m = 35.59 molal.
    Then, for 100 mL H2O = 0.100 kg,
    m x kg solvent x molar mass solute = g solute
    35.59 x 0.100 x 62.07 = 221 g ethylene glycol which is far more than you need since that would be of the order of 300 mL solution. Next, try, 50 mL H2O.

    35.59 x 0.050 x 62.07 = 110.5 g ethylene glycol or 99.6 mL. ASSUMING the volumes are additive, and I'm not sure they are, then 99.6 + 50 mL = 150 mL.

    Next try. 58.5 mL H2O
    35.59 x 0.0585 x 62.07 = 129.2 g ethylene glycol or 116.6 mL, then 116.6 + 58.5 = 175, again assuming they are additive. SO, 58.5 mL water (which you assume to be 0.0585 kg water) + 116.6 mL ethylene glycol will give you 175 mL of a solution that should boil at 118.33 IF the water and ethylene glycol are additive in volume.

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