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Preparation and mixtures of solutions

Topic: Chemistry
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Objective: Learn to count concentrations, amount of substance and volume, the relationship between these and learn to count on them. We shall also work out at the count correct number of decimal places and round the right. Introduction: By weighing the salt on an analytical balance and then bring it into a volumetric flask, pour in deionized water and turn it several times, we figure out what concentration is. We will also calculate the mass of potassium permanganate when the concentration is 0.12000 ml / dm3 in volume 100,00cm3 (ml).

Hypothesis: A hypothesis of this lab is not fitting because I have to use me the results I get from the weighted salt. I need namely to known figure in the relationship between volume, concentration, and amount of substance to be able to count on it. But I think, however, that the volume will be greater and greater, the greater concentration remains.

Material: Weighing boats made of foil, potassium permanganate, deionized water, volumetric flask, analytical, spatula and a spoon.

Performance: I'm starting to do if the volume 100,00cm3 to dm3 for the concentration unit mol / dm3. Then I know the concentration and volume, I know the two numbers in the ratio, and the only missing ingredient amount. To get subtansmängden I must multiply the concentration by volume. Amount of substance multiplied by the molar mass is mass so I must find out the molar mass of the substance KMnO4. I did this by looking at the periodic table, and add them together. There are 4 pieces O, and therefore I have to take 4 times the molar mass of the O and add to the other to get the whole substance molar mass. Then I multiplied the result molar mass of the amount of substance I substance mass. This was the task A. Task was more practical performance instead of counting. We took a foliebit and shaped like a weighing boats. We weighed only våskeppet first so that we could count how much salt is weighed. We removed the weight of the scale that we tare the scale so that we can see the importance of salt. We took initially a spatula and put the salt in vågskeppet with it was difficult with a spatula so we took a spoon instead. We waited until the value of the mass had stabilized and wrote up results. We poured this into a 100ml volumetric flask. Where you have to be very careful so as not to spill because the mass of salt will disappear without knowing how much it is because we can not count on it, and the calculation becomes less accurate. We weighed vågskeppet again to find out how much mass of salt that are stuck on vågskeppet and not come into the flask. Then we saw the wave showed that there were salt left we wrote it up as "Not loose mass of potassium permanganate". We added deionized water to about half the ball in the flask was filled. We tried to be careful that it would not get any water on the sides of the piston upper part because this will not be counted in the volume even if it's there. We touched on a little bit so we saw that potassium permanganate was solved. Since we filled with deionized water in the volumetric flask to the mark on the top of the piston. Then you have to watch so that it takes little more than the line because the water is like this:
Since the water is colored, it is difficult to see.
We put on a cap and turned the plunger up and down 50 times to be sure that the salt has dissolved properly and we get as a stock solution.

Task C: To get the stock solution volume that I should take from there to the second solution should I use the formula: Vstam C strain C = sol V lösn

If I insert the values ​​I already have, I get an equation that looks like this:
V strain 0.0020 = 0.12 0.1 And it follows that V strain = 0.002 0.1
0.12
(because 0,002 is the concentration of the first solution I have to replace the figure with the other stated concentrations when I'll calculate other volumes of stock solution)
When I want to calculate the mass of potassium permanganate will take the result of the above calculation, divided by 100 and the result multiplied by 1.8915. Thus: 2.5 cm3 1,8915g
100cm3
Since I have the volume I need to know how much there is of 100cm3 (must know the part I take). Then I know how much it is, I must take this times the mass that I have in the stock solution, partly because I have to get a unit in g (mass) and that I then find out how much mass that part volume which is the task .
Result:

Concentration mol / cm 3 V stock solution cm3 M potassium permanganate g
0.0020 1.67 0.0316
0.0030 2.5000 0.047
0.0040 3.3333 0.062
0.0050 4.167 0.078
0.0060 5.0 0.0946
0.0070 5.8333 0.11
0.0080 6.6666 0.126
0.0090 7.519802 0.142

Conclusion: The results show that my hypothesis was right, volume I took of the stock solution is increased, the larger concentrations. This is because we assume the same volume of stock solution with the same concentration to do all these solutions at different concentrations. We use deionized water in these trials, which makes it even affecting the concentration of the solution is potassium permanganate which I take from the stock solution. The results also show that the mass kailumpemanganat doubled if the concentration was doubled, for example, the concentration was 0.0020 mass 0.0316 and at 0.0040 concentration had the mass of 0.062 g. This is very logical when you need to take twice as much potassium permanganate for it to doubling the concentration (as the potassium permanganate is the only thing that gives a concentration of our efforts) The mass also increased the larger concentrations and volumes were. To a large volume must be able to demonstrate a high concentration requires the use of more mass than in a smaller volume solution with a lower concentration.

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