11:01 PM organic cc.stonybrookedu Like the pipet, the largest contributions to e

ID: 532830 • Letter: 1

Question

11:01 PM organic cc.stonybrookedu Like the pipet, the largest contributions to errors in the delivered volume are due to the user. Again, this part of the exercise is more a test of the investigators skill in the use of the buret than in establishing its intrinsic accuracy. We establish the accuracy and precision in our use of the buret by delivering two volumes of water each N25mL. We weigh the two samples and assume that the error in their weights is insignificant. The following data are collected: Sample 1 Sample 2 Units Temperature of Water 17.3 17.3 Density of Water (from Table) 0.9991 0.9991 g/mL. Buret Reading, Initial 4.86 Buret Reading, Final 32.07 27.16 mL. Initial Weight of Beaker 41.14 40.36 Final Weight of Beaker 68.52 64.8 g The Nominal volume of a sample delivered from a buret is the difference between the initial and final buret readings. That is the volume which the user observes and records and uses in subsequent calculations. Assuming the scale is properly calibrated and that no weighing errors have occurred, the Actual volume of a sample is that determined from the weight and density of the sample.

Explanation / Answer

10) The actual volume of water is determined by dividing the weight of water in the beaker by the density of water at the said temperature.

Trial 1

Trial 2

a) Initial weight of beaker (g)

41.14

40.36

b) Final weight of beaker (g)

68.52

64.8

c) Weight of water in beaker (g) = (final weight) – (initial weight) = (b) – (a)

(68.52) – (41.14) = 27.38

(64.8) – (40.36) = 24.44

d) Density of water (from table) (g/mL)

0.9991

e) Actual Volume of water (mL) = (weight of water)/(density of water) = (c)/(d)

(27.38)/(0.9991) = 27.405 27.41

(24.44)/(0.9991) = 24.462 24.46

f) Buret reading, initial (mL)

4.86

2.68

g) Buret reading, final (mL)

32.07

27.16

h) Nominal Volume of water delivered (mL) = (final buret reading) – (initial buret reading) = (g) – (f)

(32.07) – (4.86) = 27.21

(27.16) – (2.68) = 24.48

i) Difference between Actual Volume and Nominal Volume (mL) = (e) – (h)

(27.41) – (27.21) = 0.2

(24.46) – (24.48) = -0.2

11) The average of the absolute values of the deviations = ½*[(0.2) + (-0.2)] mL = 0.0 mL (ans).