Above is a lab i just did. I am struggling on putting a conclusion together wrapping it all up. Can someone help please. Doesnt have to be long just a paragraph or 2 summarizing everything.
Converting Concentrations to Different Units
Introduction:
Depending on how the concentration unit is defined, concentration can serve as a conversion between the amount of a solute and the amount of a solution or solvent. Concentrations can therefore be helpful in a number of stoichiometry issues. The original definition of the concentration unit is frequently preferred because it contains the conversion factor. Today, for chemists, it is suitable to make a solution in the lab using a specific concentration unit like molarity, and then convert the concentration to another unit. Today in the lab, I will prepare a sodium bicarbonate (baking soda) solution with a specific molarity. I will then translate the concentration into molality, mass percent, and mole fraction.
Procedure
The first step of this lab was to make sure I had everything I needed and that my surroundings were safe. I then looked at an empty beaker weighing paper and a balance. I first took a piece of weighing paper and put it on the balance, then clicked tare. I took the lid off of the baking soda (sodium bicarbonate, NaHCO3) and scooped out a gram of it, and transferred it to the balance. I repeated this step six more times so my total would come to 7 grams. I got 6.96 grams but I rounded it to 7. Once I had 7 grams of NaHCO3 I transferred it from the weighing paper on the balance to the empty beaker. Once in the beaker, I took it over to an empty 50ml cylinder. I poured the 7 grams of NaHCO3 from the beaker into the 50 ml cylinder and took the 50 ml cylinder over to the sink. I filled the cylinder until it was full and had an aqueous solution.
Results:
6.96 grams of NaHCO3 rounded to 7 grams.
End of the lab. The tallest cylinder contains 7 grams of NaHCO3 and water.
|
Mass NaHCO3 |
7 g |
|
Moles NaHCO3 |
0.0833 |
|
Liters NaHCO3 |
50x10^-3 (0.05) |
Discussion:
This lab was interesting. In order to get solid results I had to do some calculations. So the Mass of NaHCO3 was given at 7 grams. To get the moles of NaHCO3 I used the formula Moles of NaHCO3 = mass NaHCO3 / Molar mass NaHCO3= 7.0 g / 84.007 g/mol= 0.0833 mol. Next, I had to find the liters of NaHCO3. In order to get that I used the formula # of moles in NaCHO3/volume(L) which gave me the equation 0.0833/50x10^-3=1.67. Next, I had to find out the molarity of the NaHCO3 solution in units of mol/L. For that, I divided 0.0833/50x10^-3 which ended up giving me 1.67 for the molarity. For the next step in this lab, I had to calculate the molality of the solution in units of mol/kg, if the density of the solution is 1.047 g/mL. For that, I used density x volume, which in this case is 1.047x50=52.35. I subtracted that by the mass of NaHCO3 which gave me 45.35. I turned that into 45x10^-3. I then took that and divided it by 1.047 and got an answer of 1.837 mol kg ^-1. Next, I had to calculate the mass percent of sodium bicarbonate in the solution. For that, I simply divided the mass solute x 100 by the mass of the solution. Which gave me 7x100/52.35 =13.37 for the mass perfect of NHCO3 in the solution. The last step in this lab required me to calculate the mole fraction of sodium bicarbonate in the solution. I divided 0.08/by 0.045 and got a final answer of 1.85.
Conclusion:
Above is a lab i just did. I am struggling on putting a conclusion together wrapping it all up. Can someone help please. Doesnt have to be long just a paragraph or 2 summarizing everything.
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