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.
Your approach to solving the problem is generally correct so far. However, you made a small mistake in the calculation of the molecular weight (GMW) of ethylene glycol. The correct molecular weight of ethylene glycol is 62.07 g/mol, not 62.08 g/mol. Let's continue solving the equation with the corrected value:
18.33 = 571.032X / (1031.408 - 61.89376X)
To isolate X on one side of the equation, we can cross multiply and simplify:
18.33 * (1031.408 - 61.89376X) = 571.032X
Now, distribute the 18.33 to both terms in the parentheses:
18.33 * 1031.408 - 18.33 * 61.89376X = 571.032X
18918.02224 - 1132.1363888X = 571.032X
Bring the terms involving X to one side of the equation:
1132.1363888X + 571.032X = 18918.02224
Combine like terms:
1703.1683888X = 18918.02224
Divide both sides by 1703.1683888:
X = 18918.02224 / 1703.1683888
X ≈ 11.11 mL
So, the volume of ethylene glycol (C2H4(OH)2) needed is approximately 11.11 mL, and the volume of water needed would be 175 mL - 11.11 mL = 163.89 mL.