When 1 mol [CrCl3(H2O)3]6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is:
A: [CrCl3(H2O)3 ].3H2O
B: [CrCl2(H2O)4]Cl.2H2O
C: [CrCl2(H2O)5]Cl2H2O
D: [Cr(H2O)6]Cl3
When 1 mol [CrCl3(H2O)3]6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is:
A: [CrCl3(H2O)3 ].3H2O
B: [CrCl2(H2O)4]Cl.2H2O
C: [CrCl2(H2O)5]Cl2H2O
D: [Cr(H2O)6]Cl3
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1 Answer
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This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Correct option: D
When 1 mol [CrCl3 (H2O)3]6H2O is treated with excess of AgNO3, 3 mol of AgCl is produced, i.e., [CrCl3 (H2O)3]6H2Ois dissociated in aqueous solution and all three chloride comes in solution.
[Cr (H2O)6]Cl3→ [Cr (H2O)6]3+ + 3Cl-
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