   Chapter 1, Problem 62RGQ

Chapter
Section
Textbook Problem

You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 41) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm3, 11.8 g/cm3, 11.5 g/cm3, and 12.0 g/cm3. You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm3. Calculate your average value, percent error, and standard deviation of your results.

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

Percenterror=ErrorinmeasurementAcceptedvalue×100%

Error in measurement = Experimentally determined value - Accepted value.

Explanation

Average density Value:

Average density : 11.6+ 11.8 + 11.5  + 12.04                             = 11.725 g/cm3.

Hence, the average density value is obtained as shown above. Hence, the average value is 11.725 g/cm3.

Percent error:

Errorinmeasurement =11.725 g/cm3 - 11.3 g/cm3Percenterror=11.725 g/cm3 - 11.3 g/cm311.3 g/cm3×100%                       = 3.7%

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

SD =Σ|x-μ|2Nwhere,xis values in the data set,           μisaverage of the data set (mean value),            N is no.of data points.Given : Find mean value (μ):μ = 11

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