Practice Problem Set 1 l. A common approach to calibrate a piece of volumetric g
ID: 1023598 • Letter: P
Question
Practice Problem Set 1
l. A common approach to calibrate a piece of volumetric glassware is to fill the glassware with water of at a known temperature and then we the water using an analytical balance. The "true" volume e water is then calculated using the density of the water at the recorded temperature. This process is repeated several times to get an average "true" volume of water and an error in the value,
You decide to perform such an experiment on a 10.0 mL volumetric flask, filling the flask to the mark with pure Dl water at 22 C five separate times and then weighing the water added by difference using an analytical balance. You obtain the following results with the analytical balance.
Exp 1 10,0163 g
Exp 2 10.0345 g
Exp 3 10.0558 g
Exp 4 10.0292 g
Exp 5 10.0484 g
Correct these masses for the buoyancy of air (use d 0.0012 g/cm and d 8.0 g/cm and then convert the corrected masses into volumes using the density of water at22 C given in Table 2-7
What is the average volume of water contained in the volumetric glassware and the error (standard deviation) this volume?
Is the glassware marking within the expected tolerance for a Class A volumetric flask?
ls the error reflected in the 5 experimental determinations a reflection of random error or systematic error?
) What is the systematic error revealed in this experiment?
Next you weigh out 0.5516 g of KHP (potassium hydrogen phthalate) and place it in the calibrated 2. 10.0 mL volumetric flask. You then fill the flask to the mark with 30 Cwater and swirl to dissolve the KHP
a) What is the concentration (M) of the resulting KHP solution (remember to correct the mass of KHP or the buoyance of air)?
b) What is the error in the concentration of the KHP solution?
c) Which measurement/device introduces the greatest error to the concentration calculation?
3. You decide to use the solution the next day and upon measuring the temperature of the solution, find that it has decreased to 20 C
a) What is the new concentration (M) of the KHP solution and the error in this value?
b) By how much in percent did the concentration of the KHP solution change from when you made it to when you used it?
Explanation / Answer
An object’s true weight in vacum, Wv , is related to its weight in air, Wa, by the equation
Wv = Wa × [1+(1/Do1/Dw) × 0.0012]
Where Do is the object’s density at 220C = 0.999 g/cm3
Dw is the density of the calibration weight = 8.0 g/cm3
The density of air = 0.0012g/cm3
Exp 1
Wa =10.0163 g (given)
Wv = Wa × [1+(1/Do1/Dw) × 0.0012]
= 10.0163 g x [1 +( 1/0.999 g/cm3 – 1/8.0 g/cm3) x 0.0012] =10.026829g
The actual volume of water dispensed by the pipet is 10.0268g /0.999g/cm3=10.037cm3=10.037mL
If we ignore the buoyancy correction, then we report the pipet’s volume as
10.0163g /0.999g/cm3 = 10.063mL
Determinate error of (10.063mL- 10.037mL/ 10.063mL) x 100 = 0.25%.
Exp 2
10.0345 g
Wv = Wa × [1+(1/Do1/Dw) × 0.0012]
= 10.0345 g x [1 +( 1/0.999 g/cm3 – 1/8.0 g/cm3) x 0.0012] =10.045g
The actual volume of water dispensed by the pipet is 10.0345g /0.999g/cm3=10.055cm3=10.055mL
If we ignore the buoyancy correction, then we report the pipet’s volume as
10.0345g /0.999g/cm3 = 10.0445mL
Determinate error of (10.0445mL- 10.055mL/ 10.0445mL) x 100 = - 0.10%.
Exp 3
Wa = 10.0558 g
Wa × [1+(1/Do1/Dw) × 0.0012]
= 10.0558 g x [1 +( 1/0.999 g/cm3 – 1/8.0 g/cm3) x 0.0012] =10.066g
The actual volume of water dispensed by the pipet is 10.066g /0.999g/cm3=10.076cm3=10.076mL
If we ignore the buoyancy correction, then we report the pipet’s volume as
10.0558g /0.999g/cm3 = 10.0569mL
Determinate error of (10.057mL- 10.076mL/ 10.0057mL) x 100 =- 0.18%.
Exp 4
Wa= 10.0292 g
Wv = Wa × [1+(1/Do1/Dw) × 0.0012]
= 10.0292 g x [1 +( 1/0.999 g/cm3 – 1/8.0 g/cm3) x 0.0012] =10.0397g
The actual volume of water dispensed by the pipet is 10.0397g /0.999g/cm3=10.0497cm3=10.0497mL
If we ignore the buoyancy correction, then we report the pipet’s volume as
10.0292g /0.999g/cm3 = 10.0392mL
Determinate error of (10.0392mL- 10.0497mL/ 10.0392mL) x 100 = - 0.11%.
Exp 5
Wa= 10.0484 g
Wa = 10.0484 g x [1 +( 1/0.999 g/cm3 – 1/8.0 g/cm3) x 0.0012] =10.0589g
The actual volume of water dispensed by the pipet is 10.0589g /0.999g/cm3=10.0689cm3=10.0689mL
If we ignore the buoyancy correction, then we report the pipet’s volume as
10.0484g /0.999g/cm3 = 10.0584mL
Determinate error of (10.0584mL- 10.0689mL/ 10.0584mL) x 100 = - 0.10%
The average volume of water contained in the volumetric glassware sum of all volume/5
and the error (standard deviation) error in 5 experiment/5
Yes, the glassware marking within the expected tolerance for a Class A volumetric flask
ls the error reflected in the 5 experimental systematic error
There is something wrong with the instrument or its data handling system, orbecause the instrument is wrongly used by the experimenter.
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