A 0.250 g sample of a non-volatile solid dissolves in 15.0 g of tert-butanol, (F
ID: 982932 • Letter: A
Question
A 0.250 g sample of a non-volatile solid dissolves in 15.0 g of tert-butanol, (Freezing point: 25.5oC, Kf (oC m-1): 9.1). The freezing point of the solution is 20.7oC.
What is the molality of the solute in the solution?
Calculate the molar mass of the solute.
The same mass of solute, 0.250 g, is dissolved in 15.0 g of ethylene glycol, (Freezing point: -12.7oC, Kf (oC m-1): 3.11) instead of tert-butanol. What is the expected freezing point change of this solution?
Please bold your answers and show all steps for the 3 above questions.
Explanation / Answer
m = 0.25 g of solid
m = 15 g of tertbutanol
a)
molality = mol / kg solvent
dTf = -KF*m
m = dTF/Kf = (25.5-20.7)/(9.1) = 0.527472
moal = mol/kg
0.527472 = mol / 0.015
mol = 0.527472 *0.015 = 0.00791208 mol of solid
b)
dT = -KF*m
molal = 0.527472
dT = -(3.11)*0.527472 = -1.640
dTf = -1.640
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.