EXERCISE 1 ANALYSIS

# EXERCISE 1 ANALYSIS - been used to tare the scale and the...

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Bio 205L EXERCISE 1 ANALYSIS 1. For 100ml beakers: % error of mean = |(calculated ave. vol.) – 60ml| x 100% 60ml |(58.51ml) – 60ml| x 100 = 2.48% error of mean 60ml For 5ml pipets: |(4.02ml) – 4ml| x 100 = 0.5 % error of mean 4ml 2. a) 1.12 g/mL = mass mass = .28g .25mL b) 1.12 g/mL = mass mass = .056g .05mL 3. a) The results would not have been meaningful because the point of the exercises was to see which instruments were the most accurate and calibrating the automatic pipetters would have defeated that purpose. b) An analytical balance could be used to calibrate automatic pipetters accurately. 4. Since the volume of the liquid is approximately 7ml, it can be poured into a 10ml graduated cylinder in order to determine the exact volume. Since density is to be determined and the volume will be know, the liquid can be poured into a plastic taring beaker (after it has already

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Unformatted text preview: been used to tare the scale) and the mass of the liquid can be determined. Then the formula of Density = (Mass)(Volume) can be used to determine the density of the liquid. 5. density = mass/volume mass = (density)(volume) volume = (.0012 c m) 3 = 1.73 x 10-9 c m 3 density = 0.96g/cc mass = (1.73 x 10-9 μ m)(0.96g/cc) = 1.66 x 10-9 g 6. When preparing the working solutions, it was a smaller amount so a pipet was used because it needed to be more exact. However when preparing the stock solution, it was a larger quantity so a graduated cylinder could be used because there was more to transfer and it did not need to be as exact as the working solutions. 7. 0.35M = mass (58.4g)(2.0L) mass = 40.88g 8. 0.015M x 40mL = 2 mL 0.3M 9. a) 25.7g = 0.45M (342.3g)(.168L) b) .5M - .45M x 100 = 11.11% error .45M...
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EXERCISE 1 ANALYSIS - been used to tare the scale and the...

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