COLLIGATIVE PROPERTIES: FREEZING POINT DEPRESSION AND BOILING POINT ELEVATION DAY 1 – 04 FEBRUARY 2015 Colligative Properties Depends on the NUMBER of solute‚ not on the nature of solute particles Freezing Point Depression Boiling Point Elevation Vapor Pressure Lowering Osmotic Pressure Electrolyte and Nonelectrolytes Electrolytes •Separates in water forming a solution that conducts electric current •IONIC COMPOUNDS Non- electrolytes • does not allow the flow of an electric current • COVALENT
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Chemistry 121 Experiment 19 Molar Mass Determination y Depression of the Freezing Point Introduction: The most commonly used liquid is water. In this experiment we study the equilibria that can exist between pure water and an aqueous solution‚ and ice‚ the solid form of water. The heat will transfer from a higher temperature to a lower temperature. In order for water to change states of matter‚ it takes a certain amount of kinetic energy or heat. The shift from ice to water (solid to a liquid)
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Molar Mass of a Volatile Liquid There are several ways to find the molar mass of a substance. One way‚ if the substance is a gas‚ is to use the Ideal Gas Equation to find molar mass. The standard equation reads PV=nRT where “n” is the number of moles present‚ “P” is the pressure (which is obtained by reading the barometric pressure of the room with the class barometer)‚ “V” is the volume of the gas‚ “R” is the universal constant‚ and “T” is the temperature of the gas. The experiment’s objective
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Determination of the molar mass of carbon dioxide Purpose: The purpose of the practical that we completed was to determine the molar mass of carbon dioxide (CO2) by experimental means and to observe the reaction of hydrochloric acid and sodium carbonate. Using the balanced equation: Na2CO3 + 2HCl → 2NaCl + H2O + CO2 Materials / Apparatus: 1) 8.00g of sodium carbonate‚ 2) 30mL of hydrochloric acid (6molL-1) 3) 100mL of hydrochloric acid (6molL-1) 4) 100mL conical flask‚ 5) 150mL glass beaker 6)
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full determination for certain chemicals and their boiling points. It lists some already but given the atomic numbers of any material this project includes a conversion and calculation chart to find the freezing point of most any material. GOOD LUCK! Abstract: In this lab we determined the freezing point‚ and Kf‚ of pure 2‚4‚dichloralbenzne as well as a 2‚4‚dichloralbenzne/biphenyl solution. We used this information to determine the molar mass of an unknown (#24) by the 3rd step in the experiment which
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What is molar mass?Molar mass is the weight of one mole (or 6.02 x 1023 molecules) of any chemical compounds. Molar masses of common chemical compounds that you might find in the chemistry laboratory can range between 18 grams/mole for compounds like water to hundreds of grams per mole for more complex chemical compounds.The lightest possible chemical that one can have under normal conditions is hydrogen gas‚ or H2. There is no limit to how heavy a chemical compound can be - it is not uncommon for macromolecules (large
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LECTERUR : EXPERIMENT 4 DETERMINING MOLECULE WEIGHT BY FREEZING POINT DEPRESSION METHOD STUDENT NAME : ID : LAB PARTNERS | | INTRODUCTION According to Anne‚ n.d‚ the freezing point of a liquid is decreased by adding to another compound to it. This is known as freezing point depression. The pure solvent will have higher freezing point than the solution. Colligative property of matter can explain more about freezing point depression. Whereas‚ colligative properties depends on the number
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The Molar Mass of a Volatile Liquid Adam Kozdrowicz Adam Li 11/05/12 Mr. McCready Purpose: The purpose of this procedure is to determine the molar mass of an unknown liquid‚ evaporate a sample of a liquid substance‚ and measure certain physical properties of the substance as it condenses. Procedure: 1. Obtain safety goggles. 2. Trim a piece of aluminum foil so that it covers the top of a small 13 x 100 mm test tube. Secure the foil with electrical tape. Make sure
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blue color starts to fade After heating it is completely white – color has faded – and particles are smaller due to the stirring Calculations of Trial 1: Molar Mass of H2O = 2 x (1.01) + 16.00 = 18.02gmol1- How many moles of H2O evaporated Trial 1: 0.72g±0.04g/18.02gmol1- = 0.03995… ≈ 0.040mol±6.25% = 0.040mol±0.003mol Molar Mass of CuSO4 Cu: 63.55 gmol1- S=32.07 gmol1- O x 4= (16.00) x 4=64.00 gmol1- CuSO4= 159.62 gmol1- Moles of anhydrous part (CuSO4) 1.35g±0.02g/159.62 gmol1-=0
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