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Tanya [424]
3 years ago
13

The accepted value for the percent by mass of water in a hydrate is 36.0%. In a laboratory activity, a student determined the pe

rcent 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%
Chemistry
1 answer:
valentina_108 [34]3 years ago
4 0
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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The question is incomplete, the complete question is:

The solubility of slaked lime, Ca(OH)_2, in water is 0.185 g/100 ml. You will need to calculate the volume of 2.50\times 10^{-3}M HCl needed to neutralize 14.5 mL of a saturated

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Using unitary method:

In 100 mL, the mass of Ca(OH)_2 present is 0.185 g

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The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

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