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Answer :
In a solution of 38.2 mL of 0.659 M Al2(SO4)3, there are 0.050 moles of Al3+ ions and 0.075 moles of SO4 2- ions, as Al2(SO4)3 disassociates into 2 Al3+ ions and 3 SO4 2- ions in solution.
You are tasked with calculating the number of moles of Al3+ and SO42- ions in a solution of Al2(SO4)3. To begin, we need to convert the volume of the solution to liters because molarity (M) is moles/Liter. 1 L = 1000 mL, so 38.2 mL = 0.0382 L.
The molarity of Al2(SO4)3 is the number of moles of Al2(SO4)3 per liter, which means there are 0.659 moles of Al2(SO4)3 in 1 L of solution. Therefore, in 0.0382 L of solution, there are
(0.0382 L) * (0.659 moles/L) = 0.025 moles of Al2(SO4)3.
Al2(SO4)3 disassociates into 2 Al3+ ions and 3 SO42- ions in solution. Therefore, 0.025 moles of Al2(SO4)3 yields 2 * 0.025 = 0.050 moles of Al3+, and 3 * 0.025 = 0.075 moles of SO42-.
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