Aim: To determine the effect temperature and cooling rate has on crystal growth from potash solution Hypothesis: The crystal solution cooling the slowest will have the biggest crystals as the ions have time to align themselves and grow whereas solutions cooling quickly will have smaller crystals as the crystals all clump together and don’t have time to properly form. Materials: • 3 x Beakers 50ml • 1 x 1m Cooking Twine • 1 x Measuring Cylinder • 1 x Heat Mat • 1 x fridge • 1 x Centimetre Ruler • 1 x Roll of Masking Tape • 1 x Stirring Spoon • 3 x Skewer • 1 x Microscope • 1 x Scissors • 1 x Millimetre grid • 1 x Camera • 1 x Pencil • 1 x Paper • 1 x Scale Method: 1. Using a basic knot, tie one end of the string to the middle of a skewer. Once tied place the skewer at the top of the jar and let the string hang down mark where the string hits the bottom. Cut at the marking. The string was tied to the middle of a skewer using a basic knot. The skewer was placed across the beaker and the string was marked and cut where it touched the bottom of the glass. This was repeated for the other two beakers. 2. A heat pad was set up in a place free of outside interference, as was the room temperature station and a shelf was cleared in the fridge. Once setup pour 25ml of the crystal solution into each beaker. Once the solution is in mark one beaker as the room temperature, one as the fridge and one as the heat pad. 3. With the solution in the beaker the string was
During recrystallization, the solution was to be cooled to room temperature before placing it in an ice bath. Doing this allows enough time for the crystals to be formed because as the temperature decreases, the rate of crystallization slows down. If the solution was placed in the ice bath too quickly, then the cold would have blocked out the impurities and trapped them in the solution. The more impurities present, the lower the melting point so data would have been inaccurate. Also, if the melting point apparatus wasn’t set up correctly, the data would have been imprecise.
In the lab we filled the first beaker up with water. Then we took a pipet (filled with the liquid) and dropped water droplets onto the
The vial was removed from the heat and cooled to room temperature. The spin vane was rinsed with 2-3 drops of warm water over the conical vial. The vial was cooled to room temperature then placed in an ice bath for 15 minutes. The liquid was decanted from the mixture and the resulting crystals were dried on filter paper. The crystals were then placed on a watch glass for further drying. The crystals were weighed and a small sample was placed into a capillary tube for melting point determination.
The first step is the separation of the solid crystals back into separate ions, a positive ammonium ion and a negative nitrate ion. The break these ionic bonds requires a lot of energy which means heat must be taken from the surrounding water. The second step the water molecules, which are H2O, are attracted to the ions and attach themselves to the ions. The second step actually causes heat to be produced to the surrounding liquid mixture. .Even
21) After all of the solid dissolves, move the flask from the hot plate and allow it cool to room temperature. After a while, crystals should appear in the flask.
Place the beaker on the hot plate, place the thermometer in the beaker and set the hot plate to 5oC.
The first part of the lab began by one lab member adding 10.0 mL of DI water to a test tube while another lab member obtained a beaker full of ice and salt. After both these steps were complete the test tube was put in the beaker full of ice. Immediately following the test tube be being placed in the beaker, a temperature probe was inserted into the test tube. The initial temperature was recorded and after the temperature was recorded in 30 second increments. Once the water exhibited supercooling and then remained consistent at .1 °C for 3 readings it was determined that the water had froze and formed crystals. Evidence that crystals formed allowed for it to be confirmed that the water actually hit freezing point at .0
9) Trial E: Remove the syringe and empty the beaker. Add a Thermometer to the beaker. Add 200 mL of Room Temperature water to the beaker and heat with a Bunsen Burner until it reaches 100° C. Remove the Bunsen Burner. Repeat Steps 5 & 6.
Tube 4 now should only have crude solid in the tube and it is then weighed. The tube is placed into a 50℃ water bath and then approximately 0.5 -1 ml of methanol is added, as well as H2O until the solution gets cloudy, once the solution is dissolved it is cooled to room temperature and then iced. The crystals are then collected using a Hirsh funnel. Next a small amount (~ 0.1g) of the crystals are placed into a melting point tube and placed into the melting point machine to record the unknown neutral substances melting point.
As the temperature of water increases, the particles of solid Potassium chloride, KCl, which are absorbing energy from its surrounding, start moving more easily between the solution and its solid state because. According to the second law of thermodynamics, the particles will shift to the more disordered, more highly dispersed solution state. I predict that as the temperature of a KCl and water mixture increases, then the solubility of the KCl will also increase.
Submerge the graduated cylinder in the plastic tub so that it is completely filled with water. Hold the open end of the graduated cylinder and move it vertically upside-down where the open end of the graduated cylinder is still submerged in the plastic tub. Clamp the graduated cylinder the ring stand of the lab table to keep it in place. perforate a hole in the top of the rubber cork for the solution container. Cut a straw the length of about four inches. place the straw inside of the rubber cork hole. Set up your timer for two minutes.
Cut your string so that the crystal will hang in the center of the solution, and tie your string around your seed crystal, and attach the other end to U-shaped paperclip. (See figure 3)
The most common example of freezing point depression is in the salting of roads when it snows (Kimbrough, 2006). As the melting of the first flakes on the warm road occurs, a solution of salty water is created which has a lower freezing point than pure snow (Kimbrough, 2006). Now, the temperature will not get cold enough to freeze the salty solution the way it can freeze water, keeping ice and snow from bonding to the pavement (Kimbrough, 2006). In this lab, the freezing point depression was calculated experimentally by adding an unknown solute to Lauric Acid,
Crystals were collected in a Buchner funnel, washed with alcohol, then ether, then transferred into a sample tube for storage.
Maintain a minimum cooling velocity in order to avoid unfavorable mineralogical clinker phases and crystal size