Complete the last two columns given the data already present in the table.
A researcher wishes to determine how the water absorption capacity of sodium polyacrylate varies as a function of lithium ion (Li+) concentration. The researcher has recorded the following:
Concentration of LiNO3 Solution |
Mass of Sodium Polyacrylate (g) |
Incubating Volume of LiNO3 Solution (mL) |
Volume of Filtrate Collected (Unabsorbed LiNO3 Solution) (mL) |
Volume of LiNO3 Solution Absorbed by Sodium Polyacrylate (mL) |
Volume (mL) of Water Absorbed per Gram of Sodium Polyacrylate |
0.020 |
0.101 |
40.0 |
30.6 |
||
0.015 |
0.098 |
40.0 |
30.3 |
||
0.010 |
0.103 |
40.0 |
28.9 |
||
0.0050 |
0.095 |
40.0 |
27.6 |
||
0.0025 |
0.099 |
40.0 |
23.4 |
||
0.00 |
0.104 |
40.0 |
15.5 |
:
Concentration of LiNO3 Solution |
Mass of Sodium Polyacrylate (g) |
Incubating Volume of LiNO3 Solution (mL) |
Volume of Filtrate Collected (Unabsorbed LiNO3 Solution) (mL) |
Volume of LiNO3 Solution Absorbed by Sodium Polyacrylate (mL) |
Volume (mL) of Water Absorbed per Gram of Sodium Polyacrylate |
0.020 |
0.101 |
40.0 |
30.6 |
40.0-30.6=9.4 ml |
9.4 ml/0.101g=93.069 ml/g |
0.015 |
0.098 |
40.0 |
30.3 |
40.0-30.3=9.7 ml |
9.7/0.098=98.98 ml/g |
0.010 |
0.103 |
40.0 |
28.9 |
40.0-28.9=11.1ml |
11.1/0.103=107.77ml/g |
0.0050 |
0.095 |
40.0 |
27.6 |
40.0-27.6=12.4 ml |
12.4ml/0.095g=130.53ml/g |
0.0025 |
0.099 |
40.0 |
23.4 |
40.0-23.4=16.6 ml |
16.6ml/0.099 g=167.68 ml/g |
0.00 |
0.104 |
40.0 |
15.5 |
40.0-15.5=24.5 ml |
24.5ml/0.104g=235.58 ml/g |
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