Another student determined the concentration of a vinegar solution to be 0.240 mol/L. If the accepted value for the concentration of the vinegar is 0.250 mol/L, the student's percentage erreo is

Question

A. 1.00%
B. 1.24%
C. 2.40 %
D. 4.00%

D. because the difference between the two is .01. If you put .01 over .25 you will get .04 which is 4

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%error = [(0.250 - 0.240)/0.250] x 100 = ??
[(exp value - true value)/true value] x 100 = % error