You use a measuring pipet to deliver a volume of solution into a beaker. The sta
ID: 1024323 • Letter: Y
Question
You use a measuring pipet to deliver a volume of solution into a beaker. The staring reading on the pipet is 0.02 plusminus 0.02 mL and the final reading on the pipet is 8.11 plusminus 0.02 mL. Also, based on previous experiments, you know that your need to add a 0.04 plusminus 0.02 mL volume to account for the systematic error of the pipet. You then add 100.00 plusminus 0.05 mL of water to the volume of solution in the beaker. If the concentration of the initial solution was 0.2287 plusminus 0.008 M, what is the concentration of the final solution and the absolute error in this concentration? The percentage of an additive in gasoline was measured 6 times. The average plusminus standard deviation was (0.15 plusminus 0.03)% a) Calculate the range of the confidence interval at the 95% level. b) Assume that the target concentration of the additive in gasoline is 0.18%. Does the sample analyzed have a significantly different additive concentration in comparison to the target concentration at the 95% confidence level? In the laboratory you use two different approaches to measure the concentration of a substance in 6 different samples (A-F). Both methods are assumed to be subject to purely random error. a) Use the t-table to determine whether the two methods give the same result at the 95% confidence level. b) Is the standard deviation of method 1 significantly greater than the standard deviation of method 2 (95% confidence)? You perform a series of experiments in which you first generate a standard calibration curve of absorbance versus concentration by measuring the absorbance associated with 5 solutions of known concentration. Next, you measure the absorbance of an " unknown" solution 3 times. You input this information into a spreadsheet and obtain the following results: a) What is the concentration of the unknown and the error (standard deviation) in this concentration?Explanation / Answer
1.The starting reading on the pipet is 0.02 +/- 0.02 ml
The final reading on the pipet is 8.11+/- 0.02 ml.
The actual reading on the pipet is 8.11+/- 0.02 ml.- 0.02 +/- 0.02 ml
The actual reading on the pipet is 8.09 +/- 0.02 ml
Initial volume of solution = 8.09 +/- 0.02 ml
Final volume of solution = 100 +/- 0.05 ml +8.09 +/- 0.02 ml
Final volume of solution = 108.09 +/- 0.07 ml
Initial concentration of solution = 0.2287 +/- 0.0008M
Initial concentration x Initial volume of solution = final concentration x Final volume of solution
0.2287 +/- 0.0008M x 8.09 +/- 0.02 ml = final concentration x108.09 +/- 0.07 ml
0.2287 x 8.09 +/- 0.0008M x 0.02 ml = final concentration x 108.09 +/- 0.07 ml
1.8501+/- 0.000016M = final concentration x 108.09 +/- 0.07 ml
Final concentration =1.8501+/- 0.000016M/ 108.09 +/- 0.07 ml
= 0.0171 +/- 0.0002M
Percentage error in step 1 = 0.07/108.09 x 100 = 0.1880%
The percentage error in step 2 = 0.0008/0.2287 x 100 = 0.3498%
The sum of the percentage errors = 0.1880 + 0.3498 = 0.5378%
Absolute error = final concentration x percentage error/100
0.0171 x 0.5378/100 = 0.00009M
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