Question 2. Percent Relative Error Calculation (2 p). The density of pure isopro

Question

Question 2. Percent Relative Error Calculation (2 p). The density of pure isopropyl akcohol at room temperature (20 oC) is 0.7851 g/ mL. Given this and the result from Question 1, determine the percent relative error for this student's density determination To do this, first symbolically (with mathematical symbols, not actual values) state the equation to calculate the percent relative error. Define all variables. Next, give an example calculation in which actual values are plugged in and the resuits shown. Be certain all sig figs and units are correct

Explanation / Answer

relative error percentage=[(actual value-measured value)/actual value]*100

actual value of density of pure isopropyl alcohol=0.7851 g/ml

example:

Let the measured value of density of pure isopropyl alcohol=0.6521 g/ml

plugging in the given values in the above equation:

relative error percentage=[(actual value-measured value)/actual value]*100

                                      =[(0.7851-0.6521)/0.7851]*100

                                     =[0.133/0.7851]*100

                                    =0.169*100=16.940%

relative error percentage=16.940%

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