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”

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

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).

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.

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

Questions? Complete WS3