hi guys I really need some help with my homework its so confusing here it is.
5.00g of glucose,C6H12O6, is dissolved in 500g of acetic acid. What is the new freezing point and boiling point for the solution?
Kf,acetic acid = 3.90
Kb,acetic acid = 3.07
(normal freezing point for acetic acid =16.60 and boiling point = 118.5)
thanks=)
Convert 5.00g glucose to mols.
Calculate molality = mols/kg solvent(acetic acid)
Then delta T/Kb = molality and calcualte delta T and subtract from the normal freezing point.
The boiling point elevation is done the same way except use
delta T/Kb = molality. Calcualte delta T and add to normal boiling point.
Post your work if you get stuck.
ok mols = 5.00/180 =.027
molality = mols/kg solvent(acetic acid)
= .027/500 = .000054
kg solvent, not g. 500 g acetic acid = 0.500 kg. right?
right so its .027/.500 = .054?
I don't understand the delta T part though
First, you need to adjust the 0.027. You are allowed three places (5.00 g allows 3 s.f.) so I calculate
5.00/180.2 and that divided by 0.500 = 0.0555 molal.
delta T = 1.86*m
delta T = 1.86*0.0555
delta T = 0.10325 (I know that's too many places but I prefer to do my rounding ONLY once and that is at the end of the problem.)
delta T is simply the CHANGE in the freezing point. we know it will be lowered. So 16.60-0.10325=?? and round to 2 places after the decimal.
Instead of writing delta T, I could have written
normal freezing point - new freezing point = kf*m but it's just easier to write delta T, then subtract from the normal freezing point after we calculate delta T.
1.86 x .0555 = .10323
16.60 - .10323 = 16.50 for the freezing point
1.86 x .0555 = .10323
118.5 + .10323 = 118.6 for the boiling point
Wrong on both. Wrong on acetic acid because I screwed up. Wrong on the boiling point because Kb is not the same as Kf.
What did I do wrong? I used 1.86 for Kf but that is the freezing point constant for WATER. I'm so used to doing it with water that I forgot. You need to go back through but substitute the correct Kf which is given in the problem. I see it is listed in your problem as 3.90.
So delta T = 3.90*m and subtract that from 16.60. Sorry about that.
For the boiling point, use
delta T = 3.07*m and ADD to the normal boiling point of 118.5.
3.90 x .0555 = .21645
3.07 x .0555 = .170385
.21645 - 16.60 = 16.38
.170385 + 180.5 = 180.67
3.90 x .0555 = .21645
3.07 x .0555 = .170385
.21645 - 16.60 = 16.38 This should be reversed: 16.60 - 0.21645 = 16.38.
.170385 + 180.5 = 180.67 Check this number. Isn't it 118.5 for the normal boiling point? The number to be added to it, 0.17 is correct.
You made a couple of errors in your calculations. Let's go through it again step by step.
1. First, convert the mass of glucose to moles.
Moles of glucose = 5.00 g / molar mass of glucose (C6H12O6)
The molar mass of glucose (C6H12O6) is 180.2 g/mol, so:
Moles of glucose = 5.00 g / 180.2 g/mol = 0.0277 mol
2. Calculate the molality (mol/kg solvent) of the solution.
Molality = moles of solute / kg of solvent
The mass of acetic acid is given as 500 g, so we convert it to kg:
Mass of acetic acid = 500 g = 0.500 kg
Molality = 0.0277 mol / 0.500 kg = 0.0554 mol/kg
3. Calculate the change in freezing point using the equation:
ΔT = Kf * molality
The given Kf value for acetic acid is 3.90.
ΔT = 3.90 * 0.0554 mol/kg = 0.2155 °C
To find the new freezing point, subtract ΔT from the normal freezing point of acetic acid (16.60 °C):
New Freezing Point = 16.60 °C - 0.2155 °C = 16.38 °C
So the new freezing point for the solution is 16.38 °C.
4. Calculate the change in boiling point using the equation:
ΔT = Kb * molality
The given Kb value for acetic acid is 3.07.
ΔT = 3.07 * 0.0554 mol/kg = 0.1692 °C
To find the new boiling point, add ΔT to the normal boiling point of acetic acid (118.50 °C):
New Boiling Point = 118.50 °C + 0.1692 °C = 118.67 °C
So the new boiling point for the solution is 118.67 °C.
I hope this helps! Let me know if you have any further questions.