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When 29 mL of a 405 mL stock solution containing 2.01 M AlCl3 is diluted to 103 mL, how many moles of Al3+ are in the new solution?

When 29 mL of a 405 mL stock solution containing 2.01 M AlCl3 is diluted to 103 mL, how many moles of Al3+ are in the new solution?

When 29 mL of a 405 mL stock solution containing 2.01 M AlCl3 is diluted to 103mL, how many moles of Al3+ are in the new solution?Ans.According to the formula:M1V1=M2V2M1= molarity of the...
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