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6 April, 11:00

A student took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and found it read 186.5 g. What was the student's percent error?

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  1. 6 April, 11:23
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    The original or accepted value for the percent by mass of water in a hydrate = 36%

    Percen by mass of water in the hydrate determined by the student

    in the laboratory = 37.8%

    So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0) %

    = 1.8%

    So the percentage of error made by the student = (1.8/36) * 100 percent

    = (18/360) * 100 percent

    = (1/20) * 100 percent

    = 5 percent

    So the student makes an error of 5%. Option "1" is the correct option.
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