HELP!! Colligative Properties: Freezing Point Depression Experiment Procedure: 1
ID: 1013597 • Letter: H
Question
HELP!!
Colligative Properties: Freezing Point Depression Experiment
Procedure:
1. set up apparatus
2. add the stir bar to one beaker, then measure the mass of the 2
3. tare the balance, then use graduated cylinder to add 50 mL of water. record the mass.
4. add approximately the amount of sucrose (Trial 1(17.138g)), Fe(NO3)(Trial 2 (20.097g)), NaCl(Trial 3(2.932g)) and CaCl2(Trial 4(7.349g) required to make 100 mL of 0.50m solution.
5.place second beaker on balance. tare the balance, then add about 50 g of ice. record mass
6. once substance dissolves, carefully transfer the ice into the beaker containing solution. immediately lower thermometer into solution.
7.record temp
Questions:
1. calculate the molality of each solution. you will need the masses of solute, water and ice. (remember to adjust the mass of water if the solutes are hydrates)
2. Use the chemical equations for the dissolution of each solute to determine the theoretical value of i for each solute.
5. using the experimental freezing point depression calculate the experimental i for each solute
Data:
Mass (g)
Sucrose
Fe(NO3)3*9H2O
NaCl
CaCl2
Stir Bar + Beaker
71.84
71.82
71.76
71.83
Water
48.14
49.39
49.26
50.01
Weigh boat
1.704
1.905
1.905
1.905
Weigh boat + Substance
18.84
22.002
4.825
9.254
Substance
17.14
20.10
2.932
7.349
Beaker
59.44
59.45
59.44
59.46
Beaker+ Ice
109.6
110.7
109.7
109.7
Ice
50.13
51.21
50.27
50.21
Column1
Sucrose
Fe(NO3)3*9H2O
NaCl
CaCl2
Initial Temp Degrees C
1
1
1
1
Final Temp Degrees C
0
0
-1
-1.5
Freezing Point Depression Degrees C
-1
-3
-2
-2.5
Mass (g)
Sucrose
Fe(NO3)3*9H2O
NaCl
CaCl2
Stir Bar + Beaker
71.84
71.82
71.76
71.83
Water
48.14
49.39
49.26
50.01
Weigh boat
1.704
1.905
1.905
1.905
Weigh boat + Substance
18.84
22.002
4.825
9.254
Substance
17.14
20.10
2.932
7.349
Beaker
59.44
59.45
59.44
59.46
Beaker+ Ice
109.6
110.7
109.7
109.7
Ice
50.13
51.21
50.27
50.21
Column1
Sucrose
Fe(NO3)3*9H2O
NaCl
CaCl2
Initial Temp Degrees C
1
1
1
1
Final Temp Degrees C
0
0
-1
-1.5
Freezing Point Depression Degrees C
-1
-3
-2
-2.5
Explanation / Answer
1. Molality of a solution = mass of substance (g)/molar mass of substance (g/mol) x kg of solvent
For sucrose = 17.14/342.2965 x 0.04814 = 1.04 m
For Fe(NO3)3.9H2O = 20.10/403.999 x 0.04939 = 1.01 m
For NaCl = 2.932/58.44 x 0.04926 = 1.02 m
For CaCl2 = 7.349/110.98 x 0.05001 = 1.32 m
2. Theoretical value of i for,
sucrose = 1
Fe(NO3)3.9H2O = 4
NaCl = 2
CaCl2 = 3
3. Theroretical dTf,
For sucrose dTf = -1 x 1.86 x 1.04 = -1.93 oC
For Fe(NO3)3.9H2O dTf = -4 x 1.86 x 1.01 = -7.51 oC
For NaCl dTf = -2 x 1.86 x 1.02 = -3.80 oC
For CaCl2 = -3 x 1.86 x 1.32 = -16.74 oC
4. Experimental i for solutions,
Sucrose = -1/-1.86 x 1.04 = 0.51
Fe(NO3)3.9H2O = -3/-1.86 x 1.01 = 1.60
NaCl = -2/-1.86 x 1.02 = 1.05
CaCl2 = -2.5/-1.86 x 1.32 = 1.02
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