For results/conclusions Using the images attached Write 2 paragraphs 1st paragraph: Starts with Objective. Results of DeltaT and i stated2nd paragraph: sources of errors discussedThe objective is The colligative properties are the properties that undergo a change when a solute is introduced to a solvent. Freezing point depression is also a colligative property. The addition of a solute to a solvent reduces the freezing point of the resulting solution. The freezing point of the solution is always lower than the freezing point of the solvent. This difference in the freezing point between solvent and solution is called freezing point depression. The freezing point depression is directly proportional to the molality of the solution or the number of moles of solute present in the solution. Exp3: Calculation of van't Hoff factor(i) using Freezing Point DepressionData Table A(The temperature should reach atleast -10°C)Time (min)05101520#12345Data Table B(Continue recording data till the temperature is constant for atleast2 minutes at 0°C)FP solvent0°C(It should be 0°C)FP.mass of dl water#12345678910111213141516Time (min)0123456789101112131415Data Table C(Continue recording data till the temperature is constant ~ -14°C)- 11.5°Csolution(It should be -14°C)12mass of watermass of crystalline NaCl#10gTemperature (°C)-5°C-10°CTime (min)01Temperature (°C)3°C1°C0°℃1.002gTemperature (°C)5°C3°C Calculations123345678910111213141516mass solvent (H₂O) =molsolute = mass2345Find the molality of the solution (from Part C):mass solute (NaCl) = 1.002100.01molar mass solute (NaCl) -=molar massmsolution = molsolutekgsolvent67891011AT FP121314151.002g58.443g/m010.01714 mol0.01 kgsolvent - FP solution0°C-(-11.5°c)-9.6°C-10°C= 11.5°CCalculate van't Hoff factor (i):-11°C- 11.5°C58.443AT IK Msolution11.5°C=1x1.80°C Kg/mol x 2m11.5°C= 3.194 3.191.80°℃ Kg/molx2mggkgCalculate the change in temperature (FP Depression):g/mol10.01714 mol= 1.714 ~2 molFP solvent from Data Table BFP.= from Data Table CsolutionK₁ = 1.80°Ckg/molAT = from step 2m solution = from step 1

Given is an experiment about the calculation of the freezing point of a solution.To write about the…

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Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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For results/conclusions Using the images attached Write 2 paragraphs 1st paragraph: Starts with Objective. Results of DeltaT and i stated 2nd paragraph: sources of errors discussed The objective is The colligative properties are the properties that undergo a change when a solute is introduced to a solvent. Freezing point depression is also a colligative property. The addition of a solute to a solvent reduces the freezing point of the resulting solution. The freezing point of the solution is always lower than the freezing point of the solvent. This difference in the freezing point between solvent and solution is called freezing point depression. The freezing point depression is directly proportional to the molality of the solution or the number of moles of solute present in the solution.

Exp3: Calculation of van't Hoff factor(i) using Freezing Point Depression
Data Table A
(The temperature should reach atleast -10°C)
Time (min)
0
5
10
15
20
#
1
2
3
4
5
Data Table B
(Continue recording data till the temperature is constant for atleast
2 minutes at 0°C)
FP solvent
0°C
(It should be 0°C)
FP.
mass of dl water
#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Time (min)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Data Table C
(Continue recording data till the temperature is constant ~ -14°C)
- 11.5°C
solution
(It should be -14°C)
1
2
mass of water
mass of crystalline NaCl
#
10g
Temperature (°C)
-5°C
-10°C
Time (min)
0
1
Temperature (°C)
3°C
1°C
0°℃
1.002g
Temperature (°C)
5°C
3°C

Calculations
1
2
3
3
4
5
6
7
8
9
10
11
12
13
14
15
16
mass solvent (H₂O) =
molsolute = mass
2
3
4
5
Find the molality of the solution (from Part C):
mass solute (NaCl) = 1.002
10
0.01
molar mass solute (NaCl) -
=
molar mass
msolution = molsolute
kgsolvent
6
7
8
9
10
11
AT FP
12
13
14
15
1.002g
58.443g/m01
0.01714 mol
0.01 kg
solvent - FP solution
0°C-(-11.5°c)
-9.6°C
-10°C
= 11.5°C
Calculate van't Hoff factor (i):
-11°C
- 11.5°C
58.443
AT IK Msolution
11.5°C=1x1.80°C Kg/mol x 2m
11.5°C
= 3.194 3.19
1.80°℃ Kg/molx2m
g
g
kg
Calculate the change in temperature (FP Depression):
g/mol
10.01714 mol
= 1.714 ~2 mol
FP solvent from Data Table B
FP.
= from Data Table C
solution
K₁ = 1.80°Ckg/mol
AT = from step 2
m solution = from step 1

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