Please answer both parts! the answer is not 25.04, 0.08.... Here we look at the

Question

Please answer both parts! the answer is not 25.04, 0.08....

Here we look at the transfer pipet - a device constructed to deliver a precise fixed volume of liquid. A common manufacturer's specification of the precision of a 25 mL pipet is 0.03 mL. A volume of liquid delivered by the pipet should be reported as 25.00 t 0.03 mL. This is assuming, of course, that the person using the pipet uses it correctly. The largest contributions to errors in the delivered volume from a pipet are generally due to the user either in filling it to the measured marked, or failing to discharge the contents correctly. The laboratory exercise will test your skill in the use of the pipet by calculating the accuracy of the volume of water that you deliver. The following data are collected in determining the accuracy of a 25 mL pipet: Value Units 23.6 Temperature of Water Density of Water (from Table) Weight of Empty Beaker Weight of Beaker + Sample 1 Weight of Beaker + Sample 1 + Sample 2 0.99802 g/mL 50.14 74.97 100.06 We can test how accurately a person is using the pipet by calculating the volume of water they deliver with it. The calculated volume should be equal, or very close, to 25 mL if the person used the pipet correctly. The volume of a sample of water delivered from the pipet is calculated from the mass of this water, along with the density of water at room temperature. The calculated volume (from the mass and density) of Sample 1 of water delivered with the pipet is 24.88 mL. What is the calculated volume of Sample 2 in mL? Enter your answer to the correct number of significant figures.

Explanation / Answer

To calculate the mass of water in sample 2, you must subtract the weight of the beaker with both samples minus the weight of the beaker with sample 1, as follows:

Weight of sample 2 = (Weight of beaker + sample 1 + sample 2) - (Weight of beaker + sample 1)

Weight of sample 2 = 100.06 - 74.97 = 25.09 g

With the weight of sample 2 and using density of water the volume of sample 2 is obtained:

Volume of sample 2 = 25.09 g / 0.99802 g/mL = 25.14 mL

The average deviation is calculated as follows:

Xavg = (volume of sample 1 + volume of sample 2) / 2

Xavg = (24.88 + 25.14) / 2 = 25.01

avg deviation = (|Xavg - volume of sample 1| + |Xavg - volume of sample 2|) / 2

avg deviation = ( |25.01 - 24.88| + |25.01 - 25.14| ) / 2 = 0,13 = 13%

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