Question

3. A student took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and
found it read 196.5 9. What was the student's percent error?

Answer 1

Answer:

The answer is 1.71 %

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 mass = 200 g

error = 200 - 196.59 = 3.41

We have

$p(\%) = \frac{3.41}{200} \times 100 \\ = 1.705$

We have the final answer as

1.71 %

Hope this helps you