1. Mini-Unit: Colligative Properties
Day 3 - Notes
Mini-Unit: Colligative Properties
Calculating MM of an Unknown Solute Using Colligative Properties
Background: All objects that deal with colligative properties. (1) cars and antifreeze (2) ice cream making (3) ice melting (4) wood frog survives “frozen”

2. After today you will be able to…
Determine molar mass of a solute using the formulas for molality and change in freezing point/boiling point for a solution

3. If we know the mass of solute, mass of solvent, “i” value, and the new f.p. or b.p. of the solution, we can use it to calculate the molar mass of the solute dissolved in a solvent (water).

4. You too, can calculate the molar mass in 3 easy steps: 
Use ∆TB=(i)(KB)(m) or ∆TF=(i)(KF)(m) to get the molality of the solution.
Use m=mol/kg to solve for moles of solute.
Use MM = grams/mol to get molar mass.

5. Example
What is the molar mass of the solute if 53.5g of this solute is dissolved in 650g of water to produce a solution with a freezing point of -1.8°C? Assume i=2 for this solution. Step 1: ∆TF=(i)(KF)(m) ΔTF= i= KF= m=
mol
Step 2:
0.48m =
(650g x 1kg/1000g)
1.8°C
Must use kg of SOLVENT
1.8°C=(2)(1.86°C/m)(m) 2
m =
0.48mol/kg
mol =
0.31mol
1.86°C/m ?
Step 3:
53.5g MM = =
170g/mol
Must use g of SOLUTE
0.31mol

6. Questions? Complete WS3