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1.3. In determining the true volume of your pipet (calibration), how necessary was it for you to other words, if you had just taken the measured mass of water and assumed the density to differed? Present the mean and standard deviation obtained for both the rigorous and non-rigorous treatment. What is the the percent difference between these values? for the water density as a function of temperature and correction for air buoyancy? In t for the air buoyancy (a non-rigorous treatmen), how would your calibration results have
Expert Answer
In determining the true volume of pipet, it is necessary to make a rigorous correction for the water density as a function of temperature in order to achieve the accurate readings.
Rigorous treatment
The density of water at 22.2 oC = 0.997724 g/mL
The given measurements can be converted to volumes by subtracting the empty flask weight (i.e. 44.3142) and dividing by density of water (i.e. 0.997724 g/mL)
1. (54.2779 – 44.3142)/0.997724 = 9.9864 mL
2. (64.2779 – 44.3142)/0.997724 = 20.0092 mL
3. (74.1966 – 44.3142)/0.997724 = 29.9506 mL
4. (84.1377 – 44.3142)/0.997724 = 39.9143 mL
5. (94.0753 – 44.3142)/0.997724 = 49.8746 mL
The mean of given measurements (x̄) can be calculated as shown below.
x̄ = (9.9864 + 20.0092 + 29.9506 + 39.9143 + 49.8746)/5, i.e. 29.9470
The sample standard deviation for the above measurements (s) can be calculated as shown below.
s = {(9.9864 – 29.9470)2 + (20.0092 – 29.9470)2+ (29.9506 – 29.9470)2 + (39.9143 – 29.9470)2 + (49.8746 – 29.9470)2) / (5-1)}1/2, i.e. 15.7610
Non-rigorous treatment
The given measurements can be converted to volumes by subtracting the empty flask weight (i.e. 44.3142) and dividing by density of water (i.e. 1 g/mL)
1. (54.2779 – 44.3142)/1 = 9.9637
2. (64.2779 – 44.3142)/1 = 19.9637
3. (74.1966 – 44.3142)/1 = 29.8824
4. (84.1377 – 44.3142)/1 = 39.8235
5. (94.0753 – 44.3142)/1 = 49.7611
The mean of given measurements (x̄) can be calculated as shown below.
x̄ = (9.9637 + 19.9637 + 29.8824 + 39.8235 + 49.7611)/5, i.e. 29.8789
The sample standard deviation for the above measurements (s) can be calculated as shown below.
s = {(9.9637 – 29.8789)2 + (19.9637 – 29.8789)2+ (29.8824 – 29.8789)2 + (39.8235 – 29.8789)2 + (49.7611 – 29.8789)2) / (5-1)}1/2, i.e. 15.7252
The percent difference between the mean values obtained by rigorous and nonrigorous treatments
= {(29.9470 – 29.8789)/29.9470}*100, i.e. 0.2274%
The percent difference between the standard deviation values obtained by rigorous and nonrigorous treatments
= {(15.7610 – 15.7252)/15.7610}*100, i.e. 0.2271%
