The accepted value for the percent by mass of water in a hydrate is 36.0%. In a laboratory activity, a student determined the percent by mass of water in the hydrate to be 37.8%. What is the percent error for the student's measured value?
(1) 5.0% (3) 1.8%
(2) 4.8% (4) 0.05%

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

Answer:5%Explanation: