Title | Solubility Lab report |
---|---|
Course | General Chemistry Lab |
Institution | Rochester Institute of Technology |
Pages | 4 |
File Size | 111.8 KB |
File Type | |
Total Downloads | 25 |
Total Views | 137 |
Lab Report...
Title: Determine the solubility of potassium chloride, KCl, in distilled water as a function of temperature. Lab Section: Thursday 11pm Name: Rafi Karim Partners’ names: Ann Byerley Date: 09/20/2018
ABSTRACT: The objective of the solubility experiment is to determine the Solubility of KCl (in distilled water) as a function of Temperature. We will be using the following lab materials to carry out our task: scale, weigh boats, salt, distilled water, hot plate, thermometers, ice bath and test tubes. At higher temperatures the solubility of KCl increases with a decreasing gradient.
Procedure: 1) Begin by collecting all materials: scale, weigh boats, salt, distilled water, hot plate, thermometers, ice bath and test tubes. 2) Weigh out each salt on a scale using a weight boat ; record mass (g) 3) Prepare the solution by adding the correct amount of distilled water. And label each mixture from 1 to 5. 4) Place each mixture into the hot water batch. Stir the tube until each mixture is dissolved. 5) Remove the tube from the bath, stir it and let it cool. (May require a cold water bath). When a precipitate forms, record the temperature of that moment. 6) Calculate solubility for each temperature recorded. Use the formula, Solubility of KCL = (
20
)
Data and Results: Table 1: Solution preparation Mixture #1 1) 2) 3) 4) 5)
Distilled Water (mL, g) 5.00 5.00 5.00 5.00 5.00
KCl (g) 1.76 2.01 2.52 2.55 2.76
Table 2 : Solubility Data Mixture#
1) 2) 3) 4) 5)
Masss of KCl(g)
Mass of Water(g)
Plot on x-axis Temperature (°C)
1.76 2.01 2.52 2.55 2.76
5.00 5.00 5.00 5.00 5.00
23 35 56 80 D.N.E
Plot on y-axis Solubility of KCl (gKCl/100g H2O) 35.2 40.2 50.4 51.0 N/A
Temperature (°C) 60 50 40 30 20 10 0 0
10
20
30
40
50
Temperature (°C)
Figure 1: Solubility curve for KCl
60
70
80
90
Calculations: Mixture 1: 100 X Mixture 2: 100 X Mixture 3: 100 X Mixture 4: 100 X
1.76 5 2.01
= 35.2
=40.2
5 2.52
=50.4
5 2.55 5
=51.0
Discussion We did not obtain a result for the 5th mixture of KCl and this could be due to an error or a number of errors. There could have been human error while the measurement of the KCl or a system error due to zero error in the weighing balance. There might have been some impurities that got dissolved into the mixture which caused and error. The solubility of KCl increases with temperature because at higher temperatures, the average kinetic energy of the solute is higher and so they tend to dissociate more readily. But for NH3 the opposite happens as it is gaseous solute....