5.00 g of solid Al(NO3)3 was added to enough water to make a solution with a total volume of 100.0 mL. Al(NO3)3 is very soluble in water and this much solid will fully dissolve in a 100.0 mL solution. If 65 mL of water was added to the original 100.0 mL Al(NO3)3 solution described to the left in order to dilute the solution. Calculate the new diluted solution concentration, in Molarity? Assume the volumes of the 2 solutions are additive.
5.00 g of solid Al(NO3)3 was added to enough water to make a solution with a total volume of 100.0 mL. Al(NO3)3 is very soluble in water and this much solid will fully dissolve in a 100.0 mL solution. If 65 mL of water was added to the original 100.0 mL Al(NO3)3 solution described to the left in order to dilute the solution. Calculate the new diluted solution concentration, in Molarity? Assume the volumes of the 2 solutions are additive.
World of Chemistry, 3rd edition
3rd Edition
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 13A
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5.00 g of solid Al(NO3)3 was added to enough water to make a solution with a total volume of 100.0 mL. Al(NO3)3 is very soluble in water and this much solid will fully dissolve in a 100.0 mL solution.
If 65 mL of water was added to the original 100.0 mL Al(NO3)3 solution described to the left in order to dilute the solution. Calculate the new diluted solution concentration, in Molarity?
Assume the volumes of the 2 solutions are additive.
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