20. When 1 mol CrCl3.6H2O is treated with excess of AgN3 mol of AgCI a – InterviewSolution

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20. When 1 mol CrCl3.6H2O is treated with excess of AgN3 mol of AgCI are obtained. The formula of the complex is(a) [CrCl3(H20)3l.3H20 ( CCH,0)IC2H0(c) (CrCI(H,0)JC2.H20 d[Cr(H20)JCl3

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(iv) [Cr(H2O)6]Cl3 is the answer

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