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.
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter12: Solutions
Section: Chapter Questions
Problem 12.132QP
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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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