Question

question 3 of 10
a student measured the density of gold to be 18.5 g/cm³. the accepted value of the density of gold is 19.3 g/cm³. what is the percent error of the students measurement?
a. 5.92%
b. 4.15%
c. 0.60%
d. 2.11%

Explanation:

Step1: Calculate the error

Error = |Measured value - Accepted value| = |18.5 - 19.3| = 0.8

Step2: Calculate percent - error

Percent error = $\frac{\text{Error}}{\text{Accepted value}}\times100\%=\frac{0.8}{19.3}\times100\% \approx 4.15\%$

Answer:

B. 4.15%