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Weight of filter paper = 0.756 g Weight of filter paper + barium sulphate precipitate = 0.990 g Weight of barium sulphate precipitate W2 W1 =0.990-0.756 = 0.234 g %3D % Amount of barium in given solution = Ba BasO4 137.36 g 233.36 g 0.234 g 137.36 x 0.234 0.138 g 233.36 Mass of barium X-0.138 g So, Amount of barium in given solution = 0.138 g

Weight of filter paper = 0.756 g Weight of filter paper + barium sulphate precipitate = 0.990 g Weight of barium sulphate precipitate W2 W1 =0.990-0.756 = 0.234 g %3D % Amount of barium in given solution = Ba BasO4 137.36 g 233.36 g 0.234 g 137.36 x 0.234 0.138 g 233.36 Mass of barium X-0.138 g So, Amount of barium in given solution = 0.138 g

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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Experiment :estimation of the amount of baruim as barium sulphate in given solution What is the result and discussion? What is conclusion?
Weight of filter paper = 0.756 g Weight of filter paper + barium sulphate precipitate = 0.990 g Weight of barium sulphate precipitate W2-W1 0.990-0.756 0.234 g % Amount of barium in given solution = Ba BaSO4 137.36 g 233.36 g 0.234 g 137.36 X 0.234 0. 138 g 233.36 Mass of barium X-0.138 g. So, Amount of barium in given solution = 0.138 g
Transcribed Image Text:Weight of filter paper = 0.756 g Weight of filter paper + barium sulphate precipitate = 0.990 g Weight of barium sulphate precipitate W2-W1 0.990-0.756 0.234 g % Amount of barium in given solution = Ba BaSO4 137.36 g 233.36 g 0.234 g 137.36 X 0.234 0. 138 g 233.36 Mass of barium X-0.138 g. So, Amount of barium in given solution = 0.138 g
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