a student measured the density of gold to be 18.5 g/cm³. the accepted value of the density of gold is 19.3…

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?\na. 4.15%\nb. 0.60%\nc. 2.11%\nd. 5.92%

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?\na. 4.15%\nb. 0.60%\nc. 2.11%\nd. 5.92%

Answer

Explanation:

Step1: Recall percent - error formula

Percent error = $\left|\frac{\text{experimental value}-\text{accepted value}}{\text{accepted value}}\right|\times100%$

Step2: Identify values

Experimental value = $18.5\ g/cm^{3}$, Accepted value = $19.3\ g/cm^{3}$

Step3: Substitute values into formula

Percent error = $\left|\frac{18.5 - 19.3}{19.3}\right|\times100%=\left|\frac{- 0.8}{19.3}\right|\times100%$

Step4: Calculate

$\frac{0.8}{19.3}\times100% \approx 4.15%$

Answer:

A. 4.15%