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1.44 grams of a sample of marble (impure calcium carbonate) dissolved in 40 ml of 1.0 normality (N) of hydrochloric acid, then add an amount of distilled water to the resulting solution, and its volume became 100 ml, it was found that 25 ml of this resulting solution (mixture) needs 19 ml of 0.2 normality (N) NaOH solution. Calculate the normality (mole/L) of the mixture solution and then calculate the percentage of pure calcium carbonate in this sample. If you know that the molecular weight of pure calcium carbonate is 106 g/mol. *

1.44 grams of a sample of marble (impure calcium carbonate) dissolved in 40 ml of 1.0 normality (N) of hydrochloric acid, then add an amount of distilled water to the resulting solution, and its volume became 100 ml, it was found that 25 ml of this resulting solution (mixture) needs 19 ml of 0.2 normality (N) NaOH solution. Calculate the normality (mole/L) of the mixture solution and then calculate the percentage of pure calcium carbonate in this sample. If you know that the molecular weight of pure calcium carbonate is 106 g/mol. *

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
Chapter4: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 47E
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1.44 grams of a sample of marble (impure calcium carbonate) dissolved in 40 ml of 1.0 normality (N) of hydrochloric acid, then add an amount of distilled water to the resulting solution, and its volume became 100 ml, it was found that 25 ml of this resulting solution (mixture) needs 19 ml of 0.2 normality (N) NaOH solution. Calculate the normality (mole/L) of the mixture solution and then calculate the percentage of pure calcium carbonate in this sample. If you know that the molecular weight of pure calcium carbonate is 106 g/mol. *
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