The density of an alcohol solutionOriginal teacher's comment:
"This is a model report which is not perfect, but should give an idea of what type of work is expected in the IB physics program."
Editor's initial comments
A very simple little exercise (measuring density directly as mass over volume). The outcome is well known. If they could not find a table of densities, they did not look in the right place. The report has some of the expected elements, but there are many very unusual features.
- The numbering system seems to have been supplied by the teacher in a template?
- There are no figures or graphs.
- The error calculation obscures a trivial estimate with wordiness.
- It is most unusual to use the maximum deviation (residual) as the likely error.
- The use of "We... " at least sixteen times in the text, is irritating.
- The report is mostly padding. The whole thing could be done in 25% of the space.
1. AIMS
1.1. Research question
We decided to experimentally determine the density of a given 50 % alcohol solution. The density (d) is defined as the relation between its mass (m) and volume (V).
1.2 Hypothesis
We know from MAOL's tables (a data booklet for the Finnish school system) that the density of water is 1000 kgm-3 = 1 g cm-3 and that of alcohol about 800 kg m-3 = 0.8 g cm-3. We therefore make the hypothesis that the density of the given 50% solution will be about 900 kg m-3.
1.3. Variables
The variables we need to measure are the mass and volume of the given liquid. The temperature could also affect the result since liquids tend to expand in volume when it gets warmer.
2. METHODS AND TOOLS2.1. Apparatus
To find the volume of the liquid we used five 100 ml graduated measuring cylinders (see picture in appendix 1) and to find the mass we used an electronic scale. We also used a thermometer to check the temperature in the room (see 2.2.) in case comparing our results to values found by others would be needed.
2.2. Control of variables
To make sure everything had same temperature we placed the measuring cylinders and the given flask of liquid in the room where we would do the experiment the day before, and closed the curtains so the sun would not shine on them. We made sure that the cylinders would be totally empty when the experiment was done by cleaning them with acetone the day before (the small rests of acetone we assumed would evaporate during the night).
2.3. The method used
We put the cylinders empty on the electronic scale which was then zeroed, and then poured 50 ml of the solution into each of them and weighed them. The results are found in the tables in section 3.1.
3. DATA COLLECTION
3.1. Measurements and observations
Table 3.1.A. Mass measurementsCylinder Mass of 50 ml liquid (g)
1 44.03
2 45.12
3 45.20
4 44.97
5 44.67The average mass was 44.798 g The residuals, that is the difference between a measurement value and the average of them are given in table 3.1.B:
Table 3.1.B. ResidualsCylinder Mass residual(g)
1 (-)0.7682 0.322
3 0.402
4 0.172
5 (-)0.128The maximum residual, 0.768g for cylinder1 was taken to be the absolute uncertainty in the mass, Dm. The absolute uncertainty in the volume, DV, was estimated to be 1 ml. Using half the limit of reading would have given an uncertainty of 0.5 ml but the liquid surface was not quite horizontal, so we made this estimation. The temperature in the room was 21 °C , with an uncertainty of ±1°C.
3.2. Presentation of the measurement results
The volume V = 50 ml ± 1 ml and the mass m = 44.798 g ± 0.768 g. This value is not approximated here since it will be used for calculating the uncertainty in density in section 4.1.
4. DATA PROCESSING AND PRESENTATION
4.2. Data processing
Some processing of the raw data to find the uncertainties in the key variables was done in section 3.2. above. Using the formula for the density we get
d = m / V = 44.798 g/ 50 ml = 0.89596 g/ml = 0.89596 g cm-3.
The relative uncertainties are for mass Dm/m = 0.768 g /44.798 g = 0.0171436 (» 1.7 %) and for volume DV/V = 1 ml/50 ml = 0.02 (=2%). The relative uncertainty in the density, Dd/d = Dm/m + DV/V = 0.0171436 + 0.02 = 0.0371436 ( » 3.7 %). From this follows that Dd = 0.0371436 * 0.89596 g cm-3 = 0.0332791 g cm-3 » 0.03 g cm-3
4.3. Presentation of processed data.
The result we get is then that the density of the given liquid d = 0.86 g cm-3 ± 0.03 g cm-3 which in SI units gives d = 860 kgm-3 ± 30 kgm-3 .
5. CONCLUSION AND EVALUATION
5.1. Conclusions.
There are no given literature values for the density of a 50% solution of alcohol, but the value is, depending on temperature, very close to 1000 kgm-3 for pure water and 790 kgm-3 for pure ethanol. The average of these is 895 kgm-3 which means that the value 860 kgm-3 we reached appears to be a bit too low, if we compare the result we get to our original hypothesis in section 1.3.
Note: When dissolving alcohol into water, the volumes will not simply be added; 50 ml water + 50 ml alcohol gives only ca 95 ml solution. This is an example of a complication a student doing an investigation may not be aware of.
5.2. Evaluation of the procedures and results.
The electronic scale could give mass values with a precision of 0.001 g, but our Dm was much higher than that. It seemed difficult to pour exactly 50 ml of the liquid into the cylinder, and the method for using a precise volume could be improved. The volume of liquid used will also affect the mass measurements.
5.3. Suggested improvements.
The same investigation could have been done with a cheaper scale with a 0.01 g precision without significantly changing the outcome. We could use some volumetric flasks and a pipette to measure up the chosen volume with a smaller uncertainty.