4:24 PM Thu Feb 11* O 74%くロT+ :11, 2021 at 4:07 PMTechnique Report-Distillation and Boiling Point-Experiment 1Experimental determination of ethanol boiling point gives the following results:TrialB.P (°C)1275.677.176.3The value for the atmospheric pressure in the laboratory was measured to be 711 mmHg.1. Determine the corrected boiling point of ethanol. Show your work.2. Calculate the % error compared to the reported standard value for ethanol boiling pointwhich is 79.0 °C.13. Discuss two potential reasons that might result in errors in b.p. measurements (please donot include reasons such as: thermometer not working properly or any other instrumentalerrors).

Answered: Technique Report -Distillation and…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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4:24 PM Thu Feb 11
* O 74%
くロ
T
+ :
11, 2021 at 4:07 PM
Technique Report
-Distillation and Boiling Point-
Experiment 1
Experimental determination of ethanol boiling point gives the following results:
Trial
B.P (°C)
1
2
75.6
77.1
76.3
The value for the atmospheric pressure in the laboratory was measured to be 711 mmHg.
1. Determine the corrected boiling point of ethanol. Show your work.
2. Calculate the % error compared to the reported standard value for ethanol boiling point
which is 79.0 °C.
1
3. Discuss two potential reasons that might result in errors in b.p. measurements (please do
not include reasons such as: thermometer not working properly or any other instrumental
errors).

Expert Solution

Step 1 : Determining the corrected Boiling Point of ethanol

Value for Atmospheric Pressure in Laboratory = P_observed = 711 mm Hg

Standard Atmospheric pressure at sea level = $0.045 ° C p e r m m H g$

Trial	Boiling Point( $° C$ )
1	75.6
2	77.1
3	76.3

For trial 1

The Formula For Corrected Boiling Point is

$B P_{c o r r e c t e d} = B P_{o b s e r v e d} - (P_{o b s e r v e d} - 760 m m H g) * 0.045 ° C m m H g^{- 1} = 75.6 ° C - (711 - 760 m m H g) * 0.045 ° C m m H g^{- 1} = 77 . 805^{°} C T r i a l 2 B P_{c o r r e c t e d} = B P_{o b s e r v e d} - (P_{o b s e r v e d} - 760 m m H g) * 0.045 ° C m m H g^{- 1} = 77.1 ° C - (711 - 760 m m H g) * 0.045 ° C m m H g^{- 1} = 79 . 305^{°} C T r i a l 3 B P_{c o r r e c t e d} = B P_{o b s e r v e d} - (P_{o b s e r v e d} - 760 m m H g) * 0.045 ° C m m H g^{- 1} = 76.3 ° C - (711 - 760 m m H g) * 0.045 ° C m m H g^{- 1} = 78 . 505^{°} C$

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