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# Determine Solubility of KClO3 Salt.

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Introduction

UNIVERSITI TUNKU ABDUL RAHMAN UESB 1113 GENERAL LABORATORY I ATOMIC STRUCTURES & PERIODICITY BACHELOR OF SCIENCE (HONS) BIOTECHNOLOGY YEAR 1 SEMESTER 1 EXPERIMENT 9: DETERMINE SOLUBILITY OF KClO3 SALT. NAME/ID: HEW XIN HONG / 05UEB02306 LOW WEI SEONG / 05UEB02315 TUTORIAL GROUP: GROUP 4 DATE: 25TH JULY 2005 TUTOR: MR. LEE CHONG YONG Experiment 9 Title: Determine solubility of KClO3 salt. Objectives: * To determine the solubility of a given salt, KclO3 at different temperature. * State the different between a saturated solution and the original solution. * Plot a graph of solubility against temperature to determine the solubility. Theory and Background: In this experiment, we will study about solubility of solute in the solution. Solubility is meaning the maximum amount of a substance dissolved in a given volume of solvent. A solution is formed when a solid solute is begin to dissolve in a solvent. The particles of solute collide with the surface of the solid and attached on it. This process is called dissolve. While on the opposite way, is crystallization where particles of solute formed back by proper arrange. The process is affected by some factor that will introduce later. The dissolve and crystallization process is represent as below: dissolve Solute + solvent ? Solution crystallize The factor that influence solubility includes temperature, salinity, pH, dissolved organic matter (DOM), and co-solvents. And in this experiment, we will only using the temperature factor to influence the solubility, other factor will be briefly introduce. Generally, as temperature increase, solubility for solids or salts will increase which is follow the trends. But for solubility of liquids or gases, if temperature increase, the solubility is either increase or decrease depend on the solvent. ...read more.

Middle

7.20g/100mL The Solubility of KClO3 salt in 50 oC = 0.48g/5mL To convert into g/100mL, 0.48g x 20 = -------------- 5mL x 20 = 9.60g/100mL The Solubility of KClO3 salt in 50 oC = 0.87g/5mL To convert into g/100mL, 0.87g x 20 = -------------- 5mL x 20 = 17.40g/100mL The Solubility of KClO3 salt in 50 oC = 1.50g/5mL To convert into g/100mL, 1.50g x 20 = -------------- 5mL x 20 = 30.00g/100mL Discussion: In this experiment, we try to determine the solubility of KClO3 in different temperature (oC). In order to find the solubility, at starting of the experiment we have to grind finely about 2g of KClO3 salt in a mortar. We need to grind the KClO3 salt is to increase the surface area of KClO3 salt so that it will dissolve easily in distilled water. After that we transfer carefully the 5g grinded KClO3 salt into a measuring cylinder and add 10 cm of distilled water as well. When u transfers the grinded KClO3 salt into measuring cylinder, you have to be careful because u might drop some KClO3 salt at the top of the measuring cylinder. This will waste your KClO3 salt and to retrieve the salt which left at the top of the measuring cylinder, you have to move some distilled water from the bottom of the measuring cylinder to top of the measuring cylinder. This might be able to help you to retrieve the KClO3 salt which left on the top of the measuring cylinder. During experiment, we add the grinded KClO3 salt a little at a time and stir it until no more can dissolve in the distilled water. ...read more.

Conclusion

Polar solvent molecules can effectively separate the molecules of other polar substances. This happens when the positive end of a solvent molecule approaches the negative end of a solute molecule. A force of attraction then exists between the two molecules. The solute molecule is pulled into solution when the force overcomes the attractive force between the solute molecule and its neighboring solute molecule. In this situation, the main factor is temperature. So the heat that provide will help to separate the particles of the solute (KClO3 salt) and occupy the intervening spaces. Precaution steps: 1. We have considered the temperature, because a constant temperature must be maintain when obtaining the solubility of certain salt solution. 2. Using the pipette is very critical in this experiment. We have to transfer the solution as fast as possible because it might freeze and stacked at the exit of the pipette. 3. To ensure the KClO3 crystal has fully evaporate and completely dry, we must put the KClO3 crystal into the oven for about 110 oC and put it a bit longer for about 15mins. 4. Before weigh the KClO3 crystal by electrical balance, we have to put into desiccators. 5. A constant actual salt mass have to obtain to make sure that all the water inside is evaporated completely. Conclusion: 1. The solubility of KClO3 solution in 40 oC is 7.20g/100 ml. 2. The solubility of KClO3 solution in 50 oC is 9.60g/100ml. 3. The solubility of KClO3 solution in 60 oC is 17.40g/100ml. 4. The solubility of KClO3 solution in 80 oC is 30.00g/100ml. 5. Solubility is sensitive to temperature. 6. Generally, an increase in the temperature of the solution increases the solubility of a solid solute. A few solid solutes, however, are less soluble in warmer solutions. For all gases, solubility decreases as the temperature of the solution rises. ...read more.

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