L. A student calibrated a 10-mLl pipet by pipetting and weighing ten replicate a
ID: 716606 • Letter: L
Question
L. A student calibrated a 10-mLl pipet by pipetting and weighing ten replicate aliquots of distilled wat The student then converted the mass measurements to volume via the water density. The results follow Trial 6 Volume (mL) 10.002 9.993 9.984 9.996 9.989 Volume (mL) 9.983 9.991 9.990 9.988 9.999 Trial Calculate 1. the mean 2. the median 3. the range 4. the standard deviation 5. the percent relative standard deviation 6. If the student assumed (without calibrating the pipet) that the volume was exactly 10.00 mL, calculate the absolute error regarding #6, calculate the percent relative error 7, Il. When uncertainties (standard deviation, s) of the individual data points are not known, we usually simply apply the 'elementary rules' of significant figures. Report the answers of the problems below with correct significant figures. 1. 43.00408-0.40g 2. 23.45 mL +0.005 mL 3. [(4.003 g)(1x10* ug/g)1/250.0 mL 4. (35.44-28.991/28.99 100 5. log 38.54 6. log 2.3x108 7, log 0.03- 8. 10-255 IlIl Calculate the absolute standard deviations and the percent relative standard deviations for each of the following, then round the results to the correct number of significant figures. 1. 100.20 (10.08)-99.62(+0.06)+0.200 (+0.004) 2. -1.03 (+0.02) x10"+3.54 (+0.01) x10*- 3. 1.73 (t0.02) x10+ 2.44 (+0.03) x10 4. (100.34 (+0.05)+23.444 (+0.005)]+5.21(0.02)Explanation / Answer
Ans
The average of the volume
(10.002+9.993+9.984+9.996+9.989+9.983+9.991+9.990+9.988+9.999)/10 = 99.915/10 = 9.991
median is mean of middle two number
median = (9.989+9.983)/2 = 9.986
Range is diffence of highest and lowest value
10.002 - 9.983 = 0.0190
Satndard Deviation = squar root [{sumation (xi - average x)2} / (N-1)] , where Xi is sample no. and N is total number in the sample ( here it is 10 reading)
average X = 9.991
Standar Deviation = squart root [{(10.002 - 9.991)2 + (9.993-9.991)2 + (9.984-9.991)2 + (9.996-9.991)2 + (9.989-9.991)2 + (9.983-9.991)2 + (9.991-9.991)2 + (9.990-9.991)2 + (9.988-9.991)2 + (9.999-9.991)2 } /9 ] = 0.0026
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