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)?
yuh
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.
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.
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?
Thank you so much!
I really appreciate the explanation!
To find the molar mass of the solute, we can use the equation that relates the boiling point elevation (∆Tb) to the molality (m) of the solution and the elevation constant (Kb):
∆Tb = Kb × m
First, we need to calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. To find it, we need to determine the number of moles of solute and the mass of the solvent.
Given:
Mass of the solute (mSolute) = 1.50g
Volume of the solvent (Vsolvent) = 25.0 mL = 0.025 L
Density of H2O (dH2O) = 0.997 g/mL
1. Calculate the mass of the solvent:
Mass of the solvent (mSolvent) = Volume of the solvent × Density of H2O
mSolvent = 0.025 L × 0.997 g/mL
2. Calculate the moles of the solute:
Moles of the solute (nSolute) = Mass of the solute ÷ Molar mass of the solute
nSolute = 1.50g ÷ Molar mass of the solute
3. Calculate the molality:
Molality (m) = Moles of solute ÷ Mass of the solvent in kg
m = nSolute ÷ mSolvent
Now that we have the molality (m), we can calculate the molar mass of the solute.
4. Use the boiling point elevation equation to find the molar mass:
∆Tb = Kb × m
∆Tb = 100.95°C - 100.00°C (boiling point of pure water)
Kb was not given, so we need to look it up. The Kb for water is 0.512 °C/m.
We can rearrange the equation to solve for the molality (m):
m = ∆Tb ÷ Kb
5. Substitute the known values into the equation to find the molality (m):
∆Tb = 100.95°C - 100.00°C
m = ∆Tb ÷ Kb
Now, substitute the known values into the equation:
m = 0.95°C ÷ 0.512 °C/m
6. Calculate the molar mass by rearranging the equation:
Molar mass of the solute = Moles of solute ÷ Molality
Molar mass of the solute = 1.50g ÷ m
Substitute the known values into the equation:
Molar mass of the solute = 1.50g ÷ m
Finally, calculate the molar mass using the calculated molality (m) from step 5.