When 1 mol CrCl3⋅6H2O is treated with excess of AgNO3, 3 mol of AgCl – InterviewSolution

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1.	When 1 mol CrCl3⋅6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is :(i) [CrCl3 (H2O)3]⋅3H2O(ii) [CrCl2(H2O)4]Cl⋅2H2O(iii) [CrCl(H2O)5]Cl2⋅H2O(iv) [Cr(H2O)6]Cl3
Answer» (iv) [Cr(H₂O)₆]Cl₃

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