prelab for lab2[1]

# Prelab for lab2[1] - 0 1106 0 1077 0 1056 0 1077 0 1122 0 1077 0 1022 0 1077 4 = 0 0029 0 0021 0 0045 0 0055 4 = 0 0029 0 0021 0 0045 0 00554 = 0

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Pre-lab Monday, May 17, 2010 A student is asked to determine the concentration of a solution of NaOH. The Student performs four analyses of the solution and obtains concentrations of 0.1106M, 0.1058M, 0.1122M, and 0.1022M. Determine the average deviation and the relative average deviation for this student’s data. If the actual concentration of the solution is 0.1025M, calculate the absolute error and the relative error in this student’s results. x 1 =0.1106M x 2 =0.1058M x 3 =0.1122M x 4 =0.1022M _ X = (x 1 + x 2 + x 3 + x 1 )/4 =(0.1106+0.1058+0.1122+0.1022)/4 =(0.4308)/4 =0.1077M Average Deviation ( = - ) i 1Nxi x N =( .

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Unformatted text preview: - . + .- . + .- . + .- . ) 0 1106 0 1077 0 1056 0 1077 0 1122 0 1077 0 1022 0 1077 4 =( . +- . + . + . ) 0 0029 0 0021 0 0045 0 0055 4 = . + . + . + . 0 0029 0 0021 0 0045 0 00554 = . + . + . + . 0 0029 0 0021 0 0045 0 00554 = . 0 0154 = . 0 00375 Relative Average Deviation = Average deviationx × 100 Absolute Error = -x true value = .- . 0 1077 0 1025 = . 0 052 Relative Error = -x true valuetrue × value 100 = .-0 1077 . . × 0 10250 1025 100 = .-0 1077 = . . × 0 003750 1077 100 = . × 0 0348 100 = . % 3 5 . . × 0 10250 1025 100 =( . )× 0 5073 100 = . 50 73...
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## This note was uploaded on 05/17/2010 for the course CHEM 210064 taught by Professor Hastie during the Summer '10 term at York University.

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Prelab for lab2[1] - 0 1106 0 1077 0 1056 0 1077 0 1122 0 1077 0 1022 0 1077 4 = 0 0029 0 0021 0 0045 0 0055 4 = 0 0029 0 0021 0 0045 0 00554 = 0

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