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Aim: To investigate the effect of temperature on an unknown powder's
Prediction: As the temperature increases so will the solubility of the
powder. I think that the temperature and solubility will be directly
proportional - so that at temperature doubles so will the solubility
of the powder.
Hypothesis: I think this because my scientific theory says that:
Molecular solids, like sugar, and ionic solids, salts, both dissolve
in water. However, they both dissolve in different ways. The
intermolecular forces holding molecules of sugar together are quite
weak so when sugar is placed in water these bonds are broken and
individual molecules are released into solution. It takes energy to
break bonds between the molecules and it also takes energy to break
the hydrogen bonds in water. These hydrogen bonds have to be disrupted
in order to insert a sugar molecule into the substance. The energy
needed for this is produced by the forming of bonds between slightly
polar sucrose molecules and polar water molecules. This process works
so well between sugar and water that up to 800g of sugar can dissolve
in one litre of water.
The positive and negative ions in ionic solids, or salts, are held
together by the strong force of attraction between particles with
opposite charges. When a salt dissolves in water the ions are released
and become associated with the polar solvent molecules. Salts separate
from their ions when they dissolve in water.
There are several factors that affect solubility between different
compounds. These are:
* Temperature- If the solution process absorbs energy, endothermic,
then the solubility will be increased if there is a temperature
increase. If the solution
releases energy, exothermic, then solubility will decrease.
* Molecular Size- If the size or weight of the individual molecules is
solubility will be low because larger molecules are difficult to
surround with solvent molecules.
* Polarity- Generally only polar solute molecules will dissolve in
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and only non-polar solute molecules will dissolve in non-polar
solvents. Polar solute molecules have positive and negative ends. So
if a polar solute molecule is placed in a polar solvent then the
positive ends of solvent molecules will attract the negative ends of
solute molecules. This type of intermolecular force is known as di-
pole-dipole interaction. The type of intermolecular force in present
in non-polar molecules is called London Dispersion forces. Here the
positive nuclei of the atoms of the solute molecules will attract the
negative electrons of the atoms of a solvent molecule.
* Branching- This factor only applies to organic compounds. The amount
carbon branching will increase solubility because more branching will
reduce the size of the molecule, making it easier to solvate.
* Type of powder
* Amount of powder (solute)
* Amount of water (solvent)
* Solubility of solute
Make sure that you wear safety goggles as the unknown powder may be
harmful when mixed with water. At the higher temperatures be careful
with the water as you could scald yourself.
* Solvent (water)
* Solute (unknown powder)
* Water bath
* Safety goggles
* Measuring Cylinder
1. Set up a water bath and heat to 22oC (room temperature).
2. Measure out 75ml of water and 1g of powder.
3. Put the beaker of water into the water bath.
4. Wait for the temperature of solvent in the beaker to reach 30 oC.
5. Add the solute to the beaker.
6. Stir the mixture until the solute has dissolved completely.
7. Keep adding 1g of powder to the solution until the solution is
saturated; remember to stir after each gram of powder is added.
8. Record you results in a table.
9. Repeat the experiment.
10. Repeat the whole experiment for 30 oC, 40 oC, 50 oC and 60 oC
We carried out preliminary experiments to test our method, to ensure
that it works. We used different amounts of water to ensure that the
amount we had said we were going to use was appropriate. The results
of the preliminaries showed that 75 ml of water was a good amount of
water to use as the solvent as it took a reasonable amount of time for
the solution to become saturated. We also found that to make the
experiment as accurate as possible we should go up by the gram. This
means that the results are accurate to the nearest gram.
Amount of Sugar Dissolved
From looking at both my graph and my results I can see that, as the
temperature increases so does the saturation point of the solvent, and
therefore the amount of solute dissolved.
My results show a general trend that for a 10 degree rise there is a
10gram increases in the saturation point.
The reason why the amount of solute, which was sugar that dissolved as
the temperature was because the intermolecular forces holding
molecules of sugar together are quite weak so when sugar is placed in
water these bonds are broken and individual molecules are released
into solution. It takes energy to break bonds between the molecules
and it also takes energy to break the hydrogen bonds in water. These
hydrogen bonds have to be disrupted in order to insert a sugar
molecule into the substance. The energy needed for this is produced by
the forming of bonds between slightly polar sucrose molecules and
polar water molecules. Also the increase in the temperature provides
more energy to the sugar and water molecules, allowing the bonds to
break and form new bonds, more easily.
I think the results support my prediction because they show that with
the highest temperature, 60oC, the most solute dissolved, an average
of 83.0g , and with the lowest temperature, 22oC (room temperature),
the least solute dissolved, an average of 45.5g.
I think that my results are reliable enough to draw and support my
conclusion because I did the experiment at each temperature twice.
I think that the results of my experiment were reliable because I
repeated the experiment an additional time after the first time. I
repeated the experiment to make sure that my results were similar each
time and therefore reliable. I don't think that I had any anomalous
data because the results of each of the three attempts were similar
with a maximum of 3 grams between them. I felt that 3 grams was not
enough difference to repeat the experiment for a third time. The error
in the results was caused by the fact that, although we measured the
sugar to the nearest gram, the balance only measured to the nearest
0.01 gram. This means that before the experiment was carried out there
was a degree of error in the results. Also there was human error in
the reading of the results. When we were deciding if the sugar was
going to dissolve, we stirred for three minutes, and if after that
time the sugar did not dissolve then we said the solution was
saturated. This method however could be inaccurate as the sugar could
have dissolved if we had kept stirring longer. Also we were unable to
keep the way we stirred the solution constant, as we were doing it by
hand. To improve the experiment, if we were to do it a second time, we
could use automatic stirrers which would stir at the same rate and
with the same strength, so the experiment was fair. To extend the
experiment, we could use a greater range of temperatures, different
solutes and different amounts of solute and solvent.
Although the experiment gave me reliable results and I feel the method
was suitable for a classroom based experiment, I do think that it
would be necessary to make modifications to the experiment if it was
to be done on a large scale. This would be to eliminate any
inaccuracies in the measurements if the amount of sugar that dissolved
in the water, as the results were only accurate to the nearest gram.