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Sample weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL. 100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-mL of buffer solution, 50-mL of alcohol and 2.0-mL dithizone TS were added. The resulting solution required 8.7-mL of 0.031M ZnSO4 to reach the endpoint. (Note: Each mL of 0.050M EDTA is equivalent to 16.66-mg Al2(SO4)3-18H20). Compute for the percent purity of the sample.

Sample weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL. 100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-mL of buffer solution, 50-mL of alcohol and 2.0-mL dithizone TS were added. The resulting solution required 8.7-mL of 0.031M ZnSO4 to reach the endpoint. (Note: Each mL of 0.050M EDTA is equivalent to 16.66-mg Al2(SO4)3-18H20). Compute for the percent purity of the sample.

Chemistry
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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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Answer provided: 2.5% Al2(SO4)3 Please show your complete solution and write your answer clearly and readable. Thank you.
A sample of Al2(SO4)3 weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL. 100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-mL of buffer solution, 50-mL of alcohol and 2.0-mL dithizone TS were added. The resulting solution required 8.7-mL of 0.031M ZNSO, to reach the endpoint. (Note: Each mL of 0.050M EDTA is equivalent to 16.66-mg Al2(S0.)3-18H20). Compute for the percent purity of the sample.
Transcribed Image Text:A sample of Al2(SO4)3 weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL. 100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-mL of buffer solution, 50-mL of alcohol and 2.0-mL dithizone TS were added. The resulting solution required 8.7-mL of 0.031M ZNSO, to reach the endpoint. (Note: Each mL of 0.050M EDTA is equivalent to 16.66-mg Al2(S0.)3-18H20). Compute for the percent purity of the sample.
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