Boiling Point Elevation/Freezing Point Depression T = m i K Where: K b and K f d
ID: 950503 • Letter: B
Question
Boiling Point Elevation/Freezing Point Depression
T = m i K
Where:
Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.
Solvent
Formula
_Kb (° C / m)
(°
*Please note that T as defined above will be a negative number. Therefore, Kf must also be given as a negative number.
You may also see T defined as the magnitude of the freezing point depression, a positive number. In this case, K f will also be a positive number. Regardless of the sign used for K f, remember that the freezing point of the solution is always depressed.
Problem:
The boiling point of diethyl ether, CH3CH2OCH2CH3, is 34.500 °C at 1 atmosphere. Kb(diethyl ether) = 2.02 °C/m
In a laboratory experiment, students synthesized a new compound and found that when 14.98 grams of the compound were dissolved in 249.5 grams of diethyl ether, the solution began to boil at 35.173 °C. The compound was also found to be nonvolatile and a non-electrolyte.
What is the molecular weight they determined for this compound ?
__________g/mol
Explanation / Answer
T = 35.173 °C - 34.500 °C = 0.673 oC
i = 1 (the solute is a nonelectrolyte)
Kb = 2.02 °C/m
From T = m i K
m = T / iK = 0.673 oC/2.02 °C/m =0.333 molal or mol/kg
m = n mol solute / 0.2495 kg solvent
The number n of moles of solutes dissolved in 249.5 grams is
n = 0.2495 kg x 0.333 mol/kg = 0.0831 mol
Molar mass (molecular weight) = 14.98 grams / 0.0831 mol = 180.3 g/mol