Freezing Point Cyclohexane (Solvent) I. Mass of test tube and cyclohexane (grams
ID: 481065 • Letter: F
Question
Freezing Point Cyclohexane (Solvent) I. Mass of test tube and cyclohexane (grams) _________ Mass of test tube (grams) _________ Mass of cyclohexane (grams) ________ Freezing point, from cooling curve (degree C)__________ Instructor's approval of graph _____________ Freezing Point of Cyclohexane and known Solute Trial 1 Trial 2 Unknown Solute ID __________ Mass of test tube and cyclohexane (grams) ______ _______. Mass of cyclohexane (grams) _______ ________ Mass of solute, total (grams) _______ ___________ Freezing point, from cooling curve (degree C) ________ _________ Calculations k _f for cyclohexane 20.0 degree C kg/mol Freezing point change, delta Tr (degree C) _________ _________ Mass of solute in solution, total (grams) _________ _________ Mass of cyclohexane in solution (kg) _________ _________ Moles of solute (mol) _________ _________ Molar mass of solute (g/mol) _________ _________ Average molar mass of solute _________Explanation / Answer
Calculations
1. kf for cyclohexane
20.0 C.kg/mol
2. Freezing point change, Tf (C)
0.7
0.7
3. Mass of solute in solution, total (grams)
0.151
0.14
4. Mass of cyclohexane in solution (kg)
9.011*10-3 (see calculation 1 below)
7.353*10-3
Moles of solute (mol)
3.15385*10-4 (see calculation 2 below)
2.57355*10-4
Molar mass of solute (g/mol)
478.78 (see calculation 3 below)
543.99
Average molar mass of solute (g/mol)
(478.78 + 543.99)/2 = 511.385
Calculation 1: Mass of cyclohexane (g) = (9.011 g)*(1 kg/1000 g) = 9.011*10-3 kg
Calculation 2: Use the law of freezing point depression
Tf = kf. m where m = molality of the solution.
Plug in values.
0.7C = (20.0C.kg/mol)*m
===> m = 0.7/20.0 mol/kg = 0.035 mol/kg
We know that molality of a solution = moles of solute/kg of solvent.
We have
0.035 mol/kg = moles of solute/(9.011*10-3 kg)
===> moles of solute = (0.035)*(9.011*10-3) mole = 3.15385*10-4
Calculation 3: Moles of solute = mass of solute in grams/molar mass of solute in grams
We have
3.15385*10-4 mole = 0.151 g/M where M = molar mass of solute
===> M = 0.151 g/(3.15385*10-4 mole) = 478.779 g/mol 478.78 g/mol
1. kf for cyclohexane
20.0 C.kg/mol
2. Freezing point change, Tf (C)
0.7
0.7
3. Mass of solute in solution, total (grams)
0.151
0.14
4. Mass of cyclohexane in solution (kg)
9.011*10-3 (see calculation 1 below)
7.353*10-3
Moles of solute (mol)
3.15385*10-4 (see calculation 2 below)
2.57355*10-4
Molar mass of solute (g/mol)
478.78 (see calculation 3 below)
543.99
Average molar mass of solute (g/mol)
(478.78 + 543.99)/2 = 511.385
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