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.
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.
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.
It looks like you have made some progress in your calculations, but you seem to be stuck on how to isolate X in your equation. Let me guide you through the process step by step.
First, let's rewrite your equation to make it a bit easier to work with:
18.33 = 571.032X / (1031.408 - 61.89376X)
To get X by itself, you can start by multiplying both sides of the equation by (1031.408 - 61.89376X):
18.33 * (1031.408 - 61.89376X) = 571.032X
Next, distribute 18.33 to each term in the parentheses:
(18994.24264 - 1134.674116X) = 571.032X
Now, let's simplify the equation by moving all the terms containing X to one side, and the constant terms to the other side. Let's also combine like terms:
571.032X + 1134.674116X = 18994.24264
Combine the X terms on the left side:
1705.706116X = 18994.24264
Finally, to isolate X, divide both sides of the equation by 1705.706116:
X = 18994.24264 / 1705.706116
Using a calculator, you can find that X is approximately 11.14 mL.
So, to make the solution boil at 245 degrees F, you would need approximately 11.14 mL of ethylene glycol and the remaining volume of the solution (175 mL - 11.14 mL = 163.86 mL) would be water.
Please note that I assumed that the total volume of the solution is 175 mL and did not take into account the specific mass of the water or any rounding errors in the previous calculations. Make sure to double-check your numbers and adjust the solution volume accordingly.