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In an experiment, a student measured the amount of vitamin C in 200 mL of orange juice as 65 mg. The label on the carton stated the amount was 95 mg. What was the percent error in the measurement by the student? Express answer to one decimal place.

Respuesta :

The percent error in the measurement of vitamin C amount by the student is 31.6%.

PERCENTAGE ERROR:

  • The percentage error of a measurement can be calculated by dividing the difference between the exact value and the approximate value, by the exact value and then multiplied by 100. That is;

  • % error = (exact value - measured value)/exact value × 100

  • In this question, a student measured the amount of vitamin C in 200 mL of orange juice as 65 mg. The label on the carton stated the amount was 95 mg. The percent error can be calculated thus:

% error = (95 - 65)/95 × 100

% error = 30/95 × 100

% error = 31.6%.

Therefore, the percent error in the measurement of vitamin C amount by the student is 31.6%.

Learn more about percent error at: https://brainly.com/question/4170313