Question

in each of the following questions, calculate the percent error. show your work! 1) a student measured the string as 1.25 m long. the teacher said it was actually 2.12 m long. what was the students percent error? 2) the teacher said the volume of liquid was 500.0 ml. when measured, a student found it was 499.7 ml. what was the students percent error? 3) a standard mass of 250.0 g was placed on a balance. the balance said it had a mass of 243.9 grams. what is the balances percent error? 4) a teacher calculated the percent of sodium in the compound as 54%. the reference book said it was actually supposed to be 76%. what was the teachers percent error? 5) there were 34 questions on a test. even the best student in the class only got 22 correct. what was the best students percent error?

Explanation:

Step1: Recall the formula for percent error

The formula for percent error is \( \text{Percent Error}=\frac{\vert\text{Experimental Value}-\text{True Value}\vert}{\text{True Value}}\times100\%\)

Step2: Solve for problem 1

• Experimental Value \(= 1.25\) m, True Value \(=2.12\) m
• \(\vert1.25 - 2.12\vert=\vert- 0.87\vert = 0.87\)
• Percent Error\(=\frac{0.87}{2.12}\times100\%\)

\(=\frac{87}{2.12}\% \approx 41.04\%\)

Step3: Solve for problem 2

• Experimental Value \(=499.7\) mL, True Value \(=500.0\) mL
• \(\vert499.7 - 500.0\vert=\vert-0.3\vert = 0.3\)
• Percent Error\(=\frac{0.3}{500.0}\times100\%\)

\(=\frac{30}{500}\%=0.06\%\)

Step4: Solve for problem 3

• Experimental Value \(=243.9\) g, True Value \(=250.0\) g
• \(\vert243.9 - 250.0\vert=\vert-6.1\vert = 6.1\)
• Percent Error\(=\frac{6.1}{250.0}\times100\%\)

\(=\frac{610}{250}\% = 2.44\%\)

Step5: Solve for problem 4

• Experimental Value \(=54\%\), True Value \(=76\%\)
• \(\vert54 - 76\vert=\vert-22\vert = 22\)
• Percent Error\(=\frac{22}{76}\times100\%\)

\(=\frac{2200}{76}\% \approx 28.95\%\)

Step6: Solve for problem 5

• Number of wrong answers \(=34 - 22=12\)
• True Value (total questions) \(=34\), Experimental Value (wrong answers) \(=12\) (in the context of error calculation, we consider the wrong count relative to total)
• \(\vert12\vert = 12\)
• Percent Error\(=\frac{12}{34}\times100\%\)

\(=\frac{1200}{34}\% \approx 35.29\%\)

Answer:

1. Approximately \(41.04\%\)
2. \(0.06\%\)
3. \(2.44\%\)
4. Approximately \(28.95\%\)
5. Approximately \(35.29\%\)