Name:
Yang ho wang
Student ID:
Department of Chemistry
21049839
Section:
LA3
The Hong Kong University of Science and Technology
Score:
CHEM 1055 Laboratory for General Chemistry II
Lab Report 1: Colligative Properties:
Freezing Point Depression
2023-24 Spring
(To be submitted in Week 5)
[Total: 10 points]
PART I. Experimental Results
Complete the following data table.
volume of deionized water used as pure solvent = 20.0
mL (to the nearest 0.1 mL)
mass of the unknown nonelectrolyte dissolved =
g (to the nearest 0.01 g)
Time, 𝒕
(min:s)
(s)
0:00
0
0:30
30
1:00
60
1:30
90
2:00
120
2:30
150
3:00
180
3:30
210
4:00
240
4:30
270
5:00
300
5:30
330
6:00
360
6:30
390
7:00
420
7:30
450
8:00
480
8:30
510
9:00
540
9:30
570
10:00
600
Pure Solvent
Trial 1
Trial 2
Temperature, 𝑻 (°C)
(to the nearest 0.1 °C)
23.2
22.8
20.7
14.1
18.5
8.5
16.9
5.7
14.2
2.8
12.7
0.8
10.5
-0.5
9.9
-0.7
8.9
-0.1
8.2
0.0
7.5
0.0
6.9
0.0
6.5
0.0
6.1
0.0
5.6
0.0
5.0
0.0
4.4
0.0
3.6
0.0
3.0
0.0
2.4
0.0
1.9
0.0
1.00
(SampleA)
Nonelectrolyte Solution
Trial 1
Trial 2
Temperature, 𝑻 (°C)
(to the nearest 0.1 °C)
26.8
20.4
16.3
12.1
9.5
7.4
6.0
4.5
2.7
1.9
0.3
0.1
-0.9
-1.2
-1.5
-1.4
-0.8
-0.7
-0.7
-0.7
-0.7
-0.7
-0.7
-0.7
-0.7
-0.6
-0.7
-0.6
-0.7
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
-0.6
PART II. Questions
1. Use a computer to plot the cooling curves of 𝑇 (°C) against 𝑡 (s) for each trial of both pure
solvent and nonelectrolyte solution in accordance with the following requirements. [4 points]
To each cooling curve:
i. label all axes;
ii. add a linear trendline to the phase-change portion of the curve; and
iii. extend the trendline back to intersect with the curve.
Determine the freezing point (𝑇𝑓 ) (to the nearest 0.1 °C) for each trial, and average the results.
Pure Solvent (Trial 1)
Temperature vs Times
25
Temperature(°C)
20
15
10
y = -0.0001x - 0.125
R² = 0.0008
5
0
0
200
400
-5
600
800
1000
1200
1400
Times(s)
Pure Solvent (Trial 2)
Temperature vs Times
25
Temperature(°C)
20
15
y = 0.0001x - 0.0571
R² = 0.2143
10
5
0
0
100
-5
𝑇𝑓 for Trial 1 =
200
300
400
500
600
700
Times(s)
0.0
°C
𝑇𝑓 for Trial 2 =
0.0
°C
Average 𝑇𝑓 =
0.0
°C
Nonelectrolyte Solution (Trial 1)
Temperature vs Times
30
Temperature(°C)
25
20
15
y = 0.0005x - 0.8841
R² = 0.789
10
5
0
0
100
200
-5
300
400
500
600
700
Times(s)
Nonelectrolyte Solution (Trial 2)
Temperature vs Times
25
Temperature(°C)
20
15
10
y = 0.0003x - 0.7791
R² = 0.6429
5
0
0
100
-5
𝑇𝑓 for Trial 1 =
200
300
400
500
600
700
Times(s)
-0.7
°C
𝑇𝑓 for Trial 2 =
-0.7
°C
Average 𝑇𝑓 =
-0.7
°C
2. Calculate the freezing point depression ( ∆𝑇𝑓 ) caused by the addition of the unknown
nonelectrolyte to the pure solvent.
[1 point]
∆𝑇𝑓 = 0-（-0.7）
= 0.7 °C
3. Calculate the molality (𝑚) of the unknown nonelectrolyte solution.
𝑚=
[1 points]
ΔTf / i K
= 0.7/（1）（1.86）
= 0.376 mol kg-1 (to three significant figures)
4. Calculate the number of moles (𝑛) of the unknown nonelectrolyte dissolved.
(Given: density of water = 1 g mL-1)
[1 point]
𝑛 = 0.376*20/1000
= 0.00753 mol. (to three significant figures)
5. Calculate the molar mass (𝑀) of the unknown nonelectrolyte.
[1 point]
𝑀 = 1/0.00753
= 132.86 g mol-1 (to two decimal places)
6. The unknown nonelectrolyte was either glucose or sucrose. Which one was it?
[1 point]
glucose
7. Calculate the percentage error of your experimental molar mass.
[1 point]
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 = （new-old）/old
Percentage error = abs（132.86-180.156）/180.156
Percentage error = 26.2 % (to three significant figures)
***Remember to insert the image of the signed Data Sheet to Appendix on the last page.***
PART III. Appendix
Please scan/photograph the signed Data Sheet as .jpeg file and insert the file below.
(make sure your personal details, data and Staff’s/TA’s signature are clearly displayed)
Data Sheet Page 1
Data Sheet Page 2
~End~