Astudent 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?

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.

Explanation:

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