1.33 m = 1.33 moles (Et)2O in 1 kg THF.
1.33 moles = 1.33 x 74.12 grams = 98.58 g so the total solution has a mass of 1000 g + 98.58 = 1098.58 grams.
You want 471.6 grams (Et)2O which is 471.6/74.12 = 6.363 moles. So you need to take
1098.58 g solution x (6.363 moles/1.33 moles) = 5255.8 grams of the solution which rounds to 5256 grams or 5.256 kg.
A 1.33 m solution of the solute diethyl ether dissolved in the solvent tetrahydrofuran is available. Calculate the mass(kg) of the solution that must be taken to obtain 0.4716 kg of diethyl ether.
Name/Formula:
diethyl ether = (C2H5)2O
tetrahydrofuran = C4H8O
Molar Mass(g/mol):
(C2H5)2O = 74.12
C4H8O = 72.12
Density(g/mL):
(C2H5)2O = 0.7138
C4H8O = 0.8892
Answer = 5.256 kg
This question is about molality. I've tried several ways of solving this type of problem, but I never seem to come up with the correct answer. Can someone give me a step by step explanation on how to come to the answer written above?
2 answers
Thank you Dr. Bob!