Question

3 points
18) A student determines the density of gold to be 20.9g/L. The true
density of gold is 19.30g/L. What is the student's percent error?round
answer to 2 significant figures *

Answer 1

Answer:

The answer is 8.29 %

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\%$

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

$p(\%) = \frac{1.6}{19.3} \times 100 \\ = 8.290155440...$

We have the final answer as

8.29 %

Hope this helps you