A solution of 1.50g of solute dissolved in 25.0 mL of H2O at 25C has a boiling point of 100.95C. What is the molar mass of the solute if it is a nonvolatile non electrolyte and the solution behaves ideally. (d of H2O at 25C=0.997g/mL)?

Answer 1

For H2O, calculate mass.
mass = volume x density
mass = 25.0 mL x 0.997 g/mL =24.9

delta T = Kf*m
Substitute and solve for m

m = molality = #moles solute/kg solvent
kg solvent from above = 0.0249, you know mL, solve for # moles solute.

#moles solute = grams solute/molar mass solute. Solve for molar mass solute.

Answer 2

i followed your steps and got .222 and its not right. for m i calulated 13.387 and for moles of solute i got .333. and then finally i got .222. could you please help me?

Answer 3

Tori--I messed up. I guess I didn't read the question. I gave directions for the freezing point. I apologize. However, if you had followed the directions I think you would have ended up with the right answer since the only difference is in what is used for K. Kf for freezing point is 1.86 and Kb for boiling point is 0.512
Here is how it's done.

mass H2O = volume x density = 25.0*0.997 = 24.925 g
delta T = Kb*m
100.95-100 = 0.95 = 0.512*m
(Note: Use the Kb in your text/notes. 0.512 is in my notes but I often see 0.52 used.)
Solve for m and I have 1.855.
m = mol/kg solvent
1.855m = moles/0.024925
mols = 1.855 x 0.024925 = 0.04625
mol = g/molar mass so
molar mass = grams/mol = 1.50/0.04625 =
32.43 which I would round to 32.4 to 3 significant figures. You should confirm these numbers. If I round earlier I get 32.5. Let me know if you have questions. I suspect you used T (100.95) as delta T.

Answer 4

Thank you so much!
I really appreciate the explanation!

Answer 5

yuh